On the Hamilton-Jacobi Equation and Infimal Convolution in the Framework of Sobolev-functions
Hannes Luiro
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\title[Infimal Convolution in the Framework of Sobolev-functions]{On the Hamilton-Jacobi Equation and Infimal Convolution in the Framework of Sobolev-functions}
\author{Hannes Luiro}
\address{Department of Mathematics and Statistics\\
University of Jyv�skyl�\\
P.O.Box 35 (MaD)\\
40014 University of Jyv�skyl�, Finland}
\email{hannes.luiro@jyu.fi}
%\subjclass[2000]{primary 35F21, 46E35, 35B65}
%\keywords{Hamilton-Jacobi equation, Hopf-Lax formula, Sobolev spaces, regularity}
%\thanks{The author has been supported }
\maketitle
{\small \textbf{Abstract.}
We study the regularity properties of the Hamilton-Jacobi flow equation and infimal convolution in the case where initial datum function is continuous and lies in given Sobolev-space $W^{1,p}(\rn)$. We prove that under suitable assumptions it holds for solutions $w(x,t)$ that $D_xw(\cdot,t)\to Du(\cdot)$ in $L^p(\rn)$. Moreover, we construct examples showing that our results are essentially optimal.
\section{Introduction}
Consider the Hamilton-Jacobi flow equation 
\begin{equation}\label{floweq}
\begin{cases}
w_t+H(Du)=0\,\text{ in }\rn\times]0,\infty[\,,\\
w(x,0)=\lim_{t\to 0} w(x,t)=u(x)\,\text{ for every }x\in\rn\,. 
\end{cases}
\end{equation}
for initial datum function $u:\rn\to\re$ and Hamiltonian function $H:\rn\to [0,\infty[\,$.
It is known (see \cite{St}) that the unique viscosity solution $w$ of the equation (\ref{floweq}) is achieved via the Hopf-Lax formula
\begin{equation}\label{hopf}
w(x,t)=\inf_{y\in\rn}\bigg(\,u(y)+tL\big(\frac{y-x}{t}\big)\,\bigg)\,,
\end{equation}
where
\begin{equation}\label{legendre} 
L(q)=\sup_{p\in\rn}\big(\,p\cdot q-H(p)\,\big)
\end{equation}
is the Legendre-Fenchel transform of $H$. Here one of course needs to require some regularity properties for functions $u$ and $H$. It is shown in \cite{St} that (\ref{hopf}) is a solution of (\ref{floweq}) if $H$ is a strictly convex $C^1$-function with $H(0)=DH(0)=0$, and superlinear, i.e. $\frac{H(p)}{|p|}\to\infty$ as $p\to\infty$. From here on we will assume that these are valid for $H$.

For initial value function $u$ our standing assumptions are that $u$ is continuous and bounded function. In this case it is clear that $w(\cdot,t)$ provides a good approximation of $u$ when $t$ is small. Indeed, formula (\ref{hopf}) above, for given $t>0$, is also often called as an \textit{infimal convolution} of $u$ (respect to function $L$) and appears as a standard smoothing operator in the theory of viscosity solutions. In this setting the notation $w(x,t)=u_t(x)$ is typically used and so do we as well. 

We will study the properties of the infimal convolution also in the case, where $L$ needs not to be a Legendre transform of any $H$ with aforementioned properties and so that the formula (\ref{hopf}) can not be interpreted as a solution of (\ref{floweq}). Our standing assumptions for function $L$ will be that 
\begin{equation}\label{final}
L \text{ is } \text{ non-negative }C^1\text{-function},\,\, L(0)=0=DL(0) \text{ and } L \text{ is superlinear}.
\end{equation}
These assumptions almost coincide with those we made for function $H$, only the convexity assumption is dropped. It is also well known that if $L$ is the Legendre transform of $H$, then $L$ is convex function and also satisfies the required properties in (\ref{final}). 

The regularity issues for the solutions of Hamilton-Jacobi equations (or infimal convolutions) have been widely studied, see, e.g., \cite{BDR}, \cite{H}, \cite{K}, \cite{VHT} and \cite{YTW}. 
In this work we focus on the convergence problems for the derivatives of the solutions. As we know that convergence $w(\cdot, t)=u_t\to u$ holds, it is natural to ask if this convergence applies also to the derivatives. In classical setting, where $u\in C^k$, $k\in\na$ and the pointwise convergence is studied, this question is rather easy. It follows e.g. from the method of characteristics (with additional assumption $H\in C^2$) that answer is positive even for higher derivatives. 

The present paper investigates the above problem in the framework of Sobolev-functions and S. More precisely, we assume that $u$ is continuous and bounded and $u\in W^{1,p}(\rn)$ for some $1<p<\infty$ and then pose the following question: Does it hold that
\[
\tag{Q}\,\,\,\,\,Du_t\to Du\,\,\text{ in }L^p(\rn)\,\,?
\]
According to our knowledge, this problem has not been studied before.

%In this case (Q) turns out to be encapsulated around the following question: How do the possible local singularities of %Sobolev-function $u$ influence on the Sobolev-norm of $u_t$? In particular, does it hold that $\normi{Du_t}$ is uniformly %comparable to the Sobolev-norm of $u$?

%Our usual assumption will be tht $u$ is a continuous function, because this guarantees that above $w$ given by the %Hopf-Lax formula really gives the viscosity solution of (\ref{floweq}) (see [Str�mberg]). However, while in the previous %works concerning the rit is usually assumed that $u$ is continuous(at least) everywhere differentiable in classical sense %[see??]. In this work we investigate the regularity of the solutions in more general setting, i.e in the case where $u$ %is a continuous Sobolev-function. Especially, we are studying the following problem: 
%\begin{equation}\label{question}
%\text{Let } u\in W^{1,p}(\rn)\,. \text{What can we say about the convergence }\normi{D_xw\to Du}_p\,\text{ when }t\to 0\,?
%\end{equation}
%According to our knowledge, this question has not yet been studied in the previous surveys around this topic. 
%It is easy to check that assuming $u\in C^1(\rn)$ implies the above convergence. However, even in this case we have to %interpret $D_xw$ as a weak derivative since it might happen that $w(\cdot,t)$ is not differentiable at $x$, even for %arbitrarily small $t$. 
%Here it is reasonable to remark that if 
%$k\geq 2$, then $u\in C^k(\rn)$ implies that $w(\cdot,t)\in C^k(\rn)$ for every $t<t_0$ as it is shown in [?].

%Furthermore, we will prefer the notation $u_t(x):=w(x,t)$ which somehow refers to the fact that functions $u_t$ %approximate function $u$ when $t$ is small. This leads us to the another aspect of this work, the approximation of %Sobolev-functions. Indeed, we may actually forget the Hamiltonian $H$ (and equation (\ref{floweq}) as well) and choose as %a starting point the Hopf-Lax formula (\ref{hopf}) for given $L$ and then try to work out what has to be assumed on $L$ %and $u$ to guarantee that $u_t$ gives a good local Sobolev-approximation of $u$ if $t$ is small enough. This aspect also %reminds us of the important special case $L(q)=\frac{1}{2}|q|^2$, which corresponds to the so called \textit{infimal %convolution} and is typically used as a smoothing device in the theory of viscosity solutions. Below we will constantly %assume that 
%\begin{equation}\label{final}
%L \text{ is } C^1\text{-function},\,\, L(0)=0=DL(0) \text{ and } L \text{ is superlinear}.
%\end{equation}
%These assumptions coincide those for function $H$ but convexity assumption is dropped. In this setting it would be also %possible to weaken $C^1$-assumption on $L$ but this would bring technical problems which we have chosen to avoid in this %work. If $L$ is given by Legendre transformation of $H$ (with aforementioned properties for $H$), the reader will be %informed and notation $L=H^*$ may be used. It is well known that our assumptions for $H$ imply that $H^*=L$ satisfies the %properties (\ref{final}) above (and also imply that $H^*$ is strictly convex).  

%convergence results of this work can be thought to deal with the approximation of Sobolev-functions, as well. More %precisely, we may actually forget the Hamiltonian $H$ and deal with functions which are given by the  
%Hopf-Lax formula so that $L$ is only a function satisfying convenient properties.\footnote{remark that it is well known %that our assumptions for $H$ imply that its Legendre transform $L$ satisfies the same properties, especially $L$ is %strictly convex $C^1$-function, $L(0)=0=DL(0)$ and $L(x)\to\infty$ as $|x|\to\infty\,$.} From here on we will assume that %$L$ is $C^1$-function, $L(0)=0=DL(0)$ and $L(x)\to\infty$ as $|x|\to\infty\,$ (remark that here convexity is not assumed) %and the following question is considered: Does it hold that $u_t$ gives a good approximation for $u$ in a given %Sobolev-space when $t$ is small? In this setting it would be also possible to weaken $C^1$-assumption on $L$ but this %would cause technical problems which we have chosen to avoid in this work. 

%We also have to remind of the important special case $L(q)=\frac{1}{2}|q|^2=H(q)$ which corresponds to the so called %\textit{infimal convolution}, which is typically used as a smoothing device in the theory of viscosity solutions.

 
 
%Let $H:\rn\to [0,\infty]$ be continuous ('Hamiltonian')such that $H(0)=0$ ($+$ other properties). Let $u\in W^{1,p}(\rn)$.

%Here we also want to forget the Hamiltonian $H$ for a moment, thus suppose that 
%$L:\rn\to[0,\infty[$ is $C^1$-function so that $L(0)=0$ and $|L(x)|\to\infty$ as $|x|\to\infty$.
%Then suppose that $w(x,t)$ is defined by the Hopf-Lax formula above. In this work we will use the notation %$u_t(x)=w(t,x)$ which somehow indicates for the fact that functions $u_t$ are approximating functions $u$ when $t$ is %small. Another reason for this notation is the fundamental special case of the above equation, the case %$L(q)=\frac{1}{2}|q|^2$ which gives us the so called \textit{infimal convolution}. That is typically used as a smoothing %device in the theory of viscosity solutions and in this case the notation $u_t$ is typically used.  
%\[
%u_t(x)=\inf_{y\in\rn} \bigg( u(y)+tL\big(\frac{y-x}{t}\big) \bigg).
%\]

%Our problem (Q) turns out to be encapsulated around the following question: How do the possible local singularities of %the weak derivative of Sobolev-function $u$ influence on the $L^p$-norm of $Du_t$? In particular, does it hold that %$\normi{Du_t}_p$ is uniformly comparable to the Sobolev-norm of $u$?
Our first main result is the following:
\begin{theorem}\label{lptulokset}
Let $u\in W^{1,p}(\rn)\,$ be bounded, $p>n$, and suppose that there exists constant $C>0$ such that 
\begin{equation}\label{aina}
|DL(x)|\leq C\frac{L(x)}{|x|}\,.
 \end{equation}
Then  
\[\normi{Du_t-Du}_p\to 0\,\,\text{ as }t\to 0\,.\]
Moreover, $\normi{Du_t}_p\leq C(p,n,L)\normi{Du}_p$ for every $t>0$ and $u\in W^{1,p}(\rn)\,$.
\end{theorem}
It is easy to check that in the case $L(x)=C|x|^{q}$, $q>1$, appearing as the most commonly used in the theory of viscosity solutions, assumption (\ref{aina}) is valid. More careful analysis show that e.g. convexity with condition 
\begin{equation}\label{paha1}
\sup_{\frac{r}{2}<|x|<r}L\leq C\inf_{\frac{r}{2}<|x|<r}L\,
\end{equation}
for all $r>0$ and constant $C<\infty$ is sufficient for (\ref{aina}).

Regarding the solutions of the Hamilton-Jacobi equation, Theorem \ref{lptulokset}, combined with Proposition \ref{pakko} in section \ref{osa2}, gives the following Corollary:
\begin{corollary}\label{seuraus1}
Let $p>n$, $u\in W^{1,p}(\rn)$ bounded and assume that $H$ satisfies our standing assumptions (given above). Additionally, suppose that there exists $C>1$ and $C'>1$ such that 
\begin{equation}\label{paha2a}
\frac{H(2x)}{H(x)}>\,2C\text{ for every }x\in\rn\,\text{ and }
\end{equation} 
\begin{equation}\label{paha3a}
\sup_{\partial B(0,r)}\,H\,< C'\inf_{\partial B(0,r)}\,H\,.
\end{equation}
for every $r>0$. Then $D_xw(\cdot,t)\to Du(\cdot)$ in $W^{1,p}(\rn)$ if $t\to 0\,$, and $\normi{D_xw(\cdot,t)}_p\leq C(p,n,H)\normi{Du}_p$ for every $t>0$ and $u\in W^{1,p}(\rn)\,$.
\end{corollary}
\noindent The proofs of the above results are given in section \ref{osa2}.

In section \ref{osa3} we show that our assumptions in Theorem \ref{lptulokset} (and Corollary \ref{seuraus1}) are essentially sharp. 
%It is clear that in the case of inf-convolution, i.e. $L(x)=C|x|^2\,$ or more generally, when $L(x)=C|x|^{q}$, $q>1$ %assumption (\ref{aina}) is valid. More careful analysis show that e.g. convexity with condition 
%\begin{equation}\label{paha1}
%\inf_{\frac{r}{2}<|x|<r)}L\leq C\sup_{\frac{r}{2}<|x|<r}L\,
%\end{equation}
%for all $r>0$ and constant $C<\infty$ is sufficient for (\ref{aina}). Furthermore, we will verify that
%\begin{proposition}
%If $L$ is a Legendre transform of $H$ (with aforementioned properties) then (\ref{aina}) follows if there exists $C>1$ so %that
%\begin{equation}\label{paha2}
%\frac{H(2x)}{H(x)}>\,2C\text{ for every }x\in\rn\,
%\end{equation}
%and moreover, there exists $C'>1$ so that 
%\begin{equation}\label{paha3}
%\inf{B(x,C'r)}^c\, H>\,\sup_{B(x,r)}\,H\,,
%\end{equation}
%for every $r>0$.
%\end{proposition}
Relating to the assumption $p>n$, it is shown (even in the most important special case $L(q)=\frac{|q|^2}{2}$) that 
\begin{theorem}\label{ekavasta}
If $n\geq 2$, then there exists continuous and compactly supported function $u\in W^{1,n}(\rn)$ such that $|Du_t|\to \infty$ when $t\to 0$ in a set of positive measure.
\end{theorem} 
\noindent This implies that answer to (Q) is definitely negative when $1\leq p\leq n\,$. 

The rest of the section \ref{osa3} deals with assumption (\ref{aina}).
The most natural case where (\ref{aina}) fails appears when $L$ behaves in infinity like exponential function, which corresponds to the case where $H$ is of type $|x|\log(|x|)\,$ and (\ref{paha2a}) does not hold for $H$. In the second example of section three it is shown that in this case the convergence (Q) fails in general also in the case $p>n\geq 3$. In this connection we also state some open questions, especially relating to the cases $n=1$ and $n=2$.

We end up with justifying in our third example the assumption of 'quasi-radiality', i.e. property (\ref{paha3a}) by considering the example in $\re �$, where $H(x,y)=C_s|x|^{s}+C_{s'}|y|^{s'}$, $s>s'>1$. This appears as a model case for the failure of (\ref{paha3a})
%, as well as for the failure of (\ref{paha1}) for the Legendre transform $L$ of $H$ 
($H$ still satisfies other required properties). We show that convergence (Q) does not hold in this situation for any $1<p<\infty$.\\  


%More careful analysis show that e.g. convexity with condition 
%\begin{equation}\label{paha1}
%\inf_{\frac{r}{2}<|x|<r}L\leq C\sup_{\frac{r}{2}<|x|<r}L\,
%\end{equation}
%for all $r>0$ and constant $C<\infty$ is sufficient for (\ref{aina}). Furthermore, we verify that
%\begin{proposition}\label{pakko}
%If $L=H^*$ is a Legendre transform of $H$ (with aforementioned properties) then (\ref{aina}) follows if there exists %$C>1$ and $C'>1$ so that
%\begin{equation}\label{paha2}
%\frac{H(2x)}{H(x)}>\,2C\text{ for every }x\in\rn\,\text{ and }
%\end{equation} 
%\begin{equation}\label{paha3}
%\inf_{B(0,C'r)^c}\, H>\,\sup_{B(0,r)}\,H\,,
%\end{equation}
%for every $r>0$.
%\end{proposition}

%Remark that above conditions may typically fail for two different reasons: If the growth of $L$ is not polynomial %(correspondingly, growth of $H$ is kind of logarithmic, thus (\ref{paha2}) fails), or if $L$ is not kind of %'quasi-radial', i.e the values at uniformly comparable distance to the origin are not uniformly comparable %(correspondingly, $H$ does not satisfy (\ref{paha3})).
  
%The most natural case when (\ref{aina}) fails appears when $L$ behaves in infinity like exponential function, which corresponds to the case where $H$ is of type $|x|\log(|x|)\,$ and (\ref{paha2}) does not hold for $H$. In the another example of section three it is shown that in this case the convergence (Q) may fail also in the case $p>n$. 

%We end up with justifying the necessity of 'quasi-radiality', i.e. properties (\ref{paha3}) and (\ref{paha1}). We will consider the functions on $\re �$ of the form $H(x,y)=C_s|x|^{s}+C_{s'}|y|^{s'}$, $s>s'>1$, which appears as a model case for the failure of (\ref{paha3}), as well as for the failure of (\ref{paha1}) for $H^*=L$ ($H$ still satisfies other required properties). We show that convergence (Q) does not hold in this situation for any $1<p<\infty$.\\ 

\textit{Acknowledgements.}
The author would like to thank Peter Lindqvist for pointing me out the open problems studied in this paper as well as Vesa Julin, Petri Juutinen and Eero Saksman for useful discussions. 

%where $L$ ('Legendre transform of $H$') is defined by
%\[
%L(q)=\sup_{p\in\rn}\bigg(p\cdot q-H(p)\bigg)\,.
%\]

%Main theorems:
%\begin{theorem}\label{conv}
%Let $p>n$ and $u\in W^{1,p}(\rn)$. Then 
%\begin{align}
%&\normi{u_t}_{1,p}\leq C(p)\normi{u}_{1,p}\,\,,\\
%&\normi{u_t-u}_{1,p}\to 0\, \text{ when } t\to 0\,\\
%& Du_t\to Du\,\,\text{ a.e. when } t\to 0\,. 
%\end{align}
%\end{theorem}
%\begin{theorem}\label{counter}
%There exists continuous function $u\in W^{1,n}(\rn)$ such that $|Du_t|\to\infty$ a.e. when $t\to 0$.
%\end{theorem}

%I have proven (proofs have not been checked very carefully) these results in the 'standard' case $L(q)=\frac{1}{2}|q|^2$. %However, I believe that at least Theorem \ref{conv} is true in the general case as well (with natural assumptions for $%H$). After quick calculation it seemed to hold that if $u$ is $C^1$ then $Du_t\to Du$ everywhere also in the general %case(this should be well known). The case $u\in W^{1,p}(\rn)$ is much more delicate but I think that arguments from the %'standard' case with small modifications could work.





%\subsection{The case of inf-convolution}

%Let us consider the case where $H(p)=\frac{1}{2}|p|^2=L(p)\,$ above. Then $u_t$ is so called inf-convolution which %usually appears as smoothing device in theory of viscosity solutions. Its usefulness comes basically from two facts. The %first is that $u_t$ converges to $u$ uniformly as $t$ tends to zero. Another one, the deeper part, is that %inf-convolution also preserves the space of viscosity supersolutions for a wide class of partial differential equations. %More precisely...[ADD DETAILS].

%The properties of inf-convolution have been widely studied, we refer here [ADD REFERENCES].

%In the theory concerning the inf-convolutions it is usually assumed, for natural reasons that function $u$ is %Lipschitz-continuous. However, in this presentation we would like to investigate also the mapping properties of %inf-convolution as an operator acting in larger function spaces. 
%This point of view becomes quite natural at least for the following reasons: the first one is simply that inf-convolution %$u_t$ is well defined whenever $u$ is everywhere defined function. Another reason is that function $u_t(x)=:w(x,t)$ can %be interpreted as a viscosity solution to the Hamilton-Jacobi flow equation 
%\[
%\begin{cases}
%\frac{\partial}{\partial t}w+\frac{1}{2}|Dw|^2=0\,\,\text{ in }\rn\times]0,\infty[\,,\\
%w=u\,\,\text{ for } t=0\,, 
%\end{cases}
%\]
%also in the case where $u$ is not Lipschitz [WHAT IS CLEVER?...CONTINUOUS? IMPORTANT! IS THIS WELL-KNOWN?].

%One observes from the above definition that $u_t\equiv -\infty$ if $u$ decreases too rapidly in infinity when compared to %the growth rate of $L$. Here we want 
%to make things more simple and assume throughout this work that function $u$ is bounded. This appear as a quite natural %assumption in the case where our main interest is on the local regularity properties of functions $u_t$. Supporting this %aspect, observe that if $u$ is not locally bounded from below, then $u_t\equiv -\infty$.  

\section{Boundedness and convergence of the gradients}\label{osa2}

Before proving Theorem \ref{lptulokset}, we have to verify some auxiliary lemmas. Althought most of these results are rather easy and well known in optimization theory, we have not find sufficiently convenient references for our purposes. Because of that we prefer to express the complete proofs. 

Firstly, it is well known that for every fixed $t>0$ functions $u_t$ are locally Lipschitz and in our case where $u$ is assumed to be bounded, functions $u_t$ are even globally Lipschitz. More precisely, suppose that $C=\sup u$ and choose $R>0$ such that $tL(a)\geq 2 C\,$ if $|a|\geq R\,$. Then one can easily verify that $u_t$ is Lipschitz with constant
\[\sup_{a\in B(0,R)}|DL(a)|\,.\] 
Here continuity of function $u$ does not have any role.

Next we remark that the Hopf-Lax formula may be written in many different ways. In this work we will exploit the following equivalent definition:
\begin{equation}
u_t(x)=\inf_{a\in\rn}\,\big( u(x+ta)+tL(a)\,\big)\,.
\end{equation}

Furthermore, since we assumed that function $u$ is bounded and $L(x)\to\infty$ as $|x|\to\infty$, it follows that for every $t>0$ there exists $R_t>0$ such that
\begin{equation}\label{lokaalisuus}
u_t(x)=\inf_{a\in B(0,R_t)}\,\big(u(x+ta)+tL(a)\,\big)\,.
\end{equation}
for every $x\in \rn$. Assuming additional properties for $L$ or $u$ would make it possible to verify more precise estimates for $R_t$. 

 %Remark that continuity of $u$ does not play any role here. 
%From this one can quite easily prove that functions $u_t$ are globally Lipschitz with constant $C_k\leq \sup_{a\in %B(0,R_t)}|DL()|$ where Let us begin with verifying that functions $u_t$ are locally Lipschitz-functions for every $t>0$. %More precisely, suppose that $x,z\in\rn$ and $0<|x-z|<1$. Suppose that $u_t(x)>u_t(z)$ and sequence $y_k$ gives the value $u_t(z)$. Then we simply estimate the difference $u_t(x)-u_t(z)$ by choosing the same sequence at point $z$, thus
%\begin{align*}
%u_t(x)-u_t(z)&=u_t(x)-\big(\lim_{k\to\infty}\, u(y_k)+tL\big(\frac{z-y_k}{t}\big)\,\big)\\&\leq 
%\lim_{k\to\infty}\,\big(\, u(y_k)+tL\big(\frac{x-y_k}{t}\big)\,\big)
%-\lim_{k\to\infty}\,\big(\, u(y_k)+tL\big(\frac{z-y_k}{t}\big)\,\big)\\
%&=t\,\lim_{k\to\infty}\big(L\big(\frac{x-y_k}{t}\big)-L\big(\frac{z-y_k}{t}\big)\big)\,.
%\end{align*}
%Then it follows from (\ref{lokaalisuus}) that $\frac{|z-y_k|}{t}\leq\frac{R}{t}$ when $k$ is big enough, thus %$\frac{|x-y_k|}{t}\leq \frac{R}{t}+\frac{1}{t}$ and we get from above that   
%\begin{align*}
% u_t(x)-u_t(z)&\leq \lim_{k\to\infty}\, %t\,\big|\frac{x-y_k}{t}-\frac{z-y_k}{t}\big|\bigg(\,\sup_{B(0,\frac{R}{t}+\frac{1}{t})} |DL|\,\bigg)\\
%&=|x-z|\bigg(\,\sup_{B(0,\frac{R}{t}+\frac{1}{t})} |DL|\,\bigg)\,.
%\end{align*} 

Throughout this work the following notation is used:
For every $x\in\rn$ define set $\R u_{t}(x)$ such that 
$a\in \R u_{t}(x)$ if 
%and only if there exists sequence of points $a_k$ converging to $a$ so that
%\[
%u_{t}(x)=\lim_{k\to\infty}\,u(x+a_kt)+tL(a_k)\big)\,.
%\]
%It is easy to check that $R u_t(x)\neq \emptyset$ (that follows for example from (\ref{lokaalisuus})). Moreover
%If $u$ is continuous then it also clearly holds that 
\begin{equation}\label{loppu2}
u_{t}(x)=u(x+at)+tL(a)\,.%\text{ for every }a\in\R u_{t}(x)\,
\end{equation}
It is easy to check that $\R u_t(x)$ is always non-empty and compact set.
%but for general $u$ this of course may not hold.
The following inequality (suppose $|u|\leq C$) 
\begin{equation}\label{loppu1}
 u_t(x)=u(x+ta_t)+tL(a_t)\geq -C+t|a_t|\frac{L(a_t)}{|a_t|},
\end{equation}
if $a_t\in \R u_t(x)$, combined with the superlinearity of $L$ (and the fact that $a_t$ minimizes (\ref{loppu2})), guarantees that for any given $x$ the sets $t\R u_t(x)$, $t>0$, are uniformly bounded (respect to $t$). This, combined with the continuity of $u$, also implies the (pointwise) convergence $u_t\to u$ as $t\to 0$.

Since functions $u_t$ are Lipschitz, they are differentiable almost everywhere by Rademacher's theorem. Actually more careful analysis show that they are even twice differentiable a.e. This is due to the fact that one can show that functions $u_t$ are semiconcave (see e.g. \cite{JJ}).

It is easy to see that functions $u_t$ do not need to be differentiable everywhere. However, the directional derivatives exist at every point, which is verified in the following lemma.  

\begin{lemma}\label{directional}
Suppose that $u$ is continuous and bounded function and $t>0$. Then $u_t$ has directional derivatives at every point to any direction $\gamma$ and
\begin{equation}
D_{\gamma}u_t(x)=\inf_{a'\in\R u_t(x)}(-D_{\gamma}L(a'))=-D_{\gamma}L(a)\,\textit{ for some }a\in\R u_t(x)\,.
\end{equation}
\end{lemma}
\textit{Proof.} Let $t>0$ and $x\in\rn$. Recall first that by compactness of set $\R u_t(x)$ there exists $a\in\R u_t(x)$ such that 
\begin{equation}
-D_{\gamma}L(a)=\inf_{a'\in\R u_t(x)}(-D_{\gamma}L(a'))\,.
\end{equation}
Moreover, continuity of $u$ implies that $u_t(x)=u(x+ta)+tL(a)$.
We first estimate the difference $u_t(x+h\gamma)-u_t(x)$, $h>0$,  from above by choosing points $a_h$ such that $a_h=a-\frac{h\gamma}{t}$. It follows that
\begin{align*}
\frac{u_t(x+h\gamma)-u_t(x)}{h}&\leq
\frac{1}{h}\big(\,(u(x+h\gamma +ta_h)+tL(a_h))-(u(x+ta)+tL(a))\,\big)\\
&=\frac{t}{h}\big(\,L(a_h)-L(a)\,\big)=\frac{\,L(a-\frac{h\gamma}{t})-L(a)}{\frac{h}{t}}\,.
\end{align*}
Taking the limit yields that
\begin{equation}
\limsup_{h\to 0^+}\,\frac{u_t(x+h\gamma)-u_t(x)}{h}\leq -D_{\gamma}L(a)\,.
\end{equation}

Assume then that there exists sequence $h_k>0$, $h_k\to 0$ as $k\to\infty$, and $\lambda >0$ so that  
\begin{equation}\label{AT}
\frac{u_t(x+h_k\gamma)-u_t(x)}{h_k}< -D_{\gamma}L(a)-\lambda\, 
\end{equation}
for every $k$. Suppose that $a_k\in \R u_t(x+h_k)$ and choose $a'_k=a_k+\frac{h_k\gamma}{t}\,$. Recall that sequence ($a_k$) has to be bounded and by extracting a subsequence, if needed, we may assume that $a_k\to a_0$. Moreover, 
\[
u_t(x)=\lim_{k\to\infty} u_t(x+h_k)=\lim_{k\to\infty}\,u(x+h_k+ta_k)+tL(a_k)=\lim_{k\to\infty}\,u(x+ta'_k)+tL(a'_k)
\]
which guarantees that $a_0\in\R u_t(x)\,$. Then we estimate the difference from below using points $a'_k$: 
\begin{align*}
\frac{u_t(x+h_k\gamma)-u_t(x)}{h_k}&\geq
\frac{1}{h_k}\big(\,(u(x+h_k\gamma +ta_k)+tL(a_k))-(u(x+ta'_k)+tL(a'_k))\,\big)\\
&=\frac{t}{h_k}\big(\,L(a_k)-L(a'_k)\,\big)=\frac{\big(\,L(a_k)-L(a_k+\frac{h_k\gamma}{t})\,\big)}{\frac{h_k}{t}}\,. 
\end{align*}
Then recall that $L$ is $C^1$-function and $a_k\to a_0$ as $k\to\infty$. This implies that above the last quantity converges to $-D_{\gamma}L(a_0)$. Because $a_0\in \R u_t(x)$ we get that
\begin{align*}
\liminf_{k\to\infty}\frac{u_t(x+h_k\gamma)-u_t(x)}{h_k}\geq -D_{\gamma}L(a_0)\geq -D_{\gamma}L(a).
\end{align*}
This contradicts with (\ref{AT}) and the proof is complete.
\hfill$\Box$\\ 
 
We get the following corollary in the points of differentiability of $u_t\,$:
\begin{corollary}\label{peruskaava}
Let $u$ be continuous and bounded and $t>0$. Then $u_t$ is differentiable at $x$ and $Du_t(x)=-DL(a)$ for every $a\in \R u_t(x)$ if and only if  
$DL(a)=DL(a')$ for every $a,a'\in\R u_t(x)\,$.
\end{corollary}
\textit{Proof.}
Suppose, on the countrary that $u_t$ is differentiable at $x$ such that there exists $a_1,a_2\in \R u_t(x)$ so that $DL(a_1)\neq DL(a_2)\,$. Then there exists direction $\gamma$ so that $DL(a_1)\cdot\gamma < DL(a_2)\cdot\gamma\,$. Then we get by Lemma (\ref{directional})  that $D_{\gamma}u_t(x)\leq -DL(a_2)\cdot\gamma<-DL(a_1)\cdot\gamma=DL(a_1)\cdot(-\gamma)\leq-D_{-\gamma}u_t(x)\,=D_{\gamma}u_t(x)\,$, which is a contradiction. Another direction follows by the same argument. 
\hfill$\Box\,$\\

\textit{Remark 1.} Though we know that functions $u_t$ are differentiable almost everywhere, we prefer writing our lemmas or theorems using directional derivatives $D_{\gamma}u_t$ since they really exist at every point for \textit{every} $t\in(0,1)$. Then we can avoid possible technical problems which may occur if there were some exceptional sets of measure zero for every $t\in(0,1)\,$. 

\textit{Remark 2.}
It is often preferable to study the regularity of the solutions $u_t$ in terms of subdifferentials (see e.g. \cite{PR}). 
In this presentation we have chosen to avoid this since they seem not to give any advantage for our short-term goals. 

\textit{Remark 3.} In above results it was assumed that $u$ is continuous function. However, this is actually not needed but simplifies the proof since it makes possible to write $u_t(x)=u(x+ta)+tL(a)$ if $a\in\R u_t(x)$ instead of using the sequences $(a_k)$ giving the infimum. 

\textit{Remark 4.} It is clear that also the boundedness of $u$ can be replaced by assuming only that the growth of $u$ in infinity is suitably comparable to the growth of $L$. This remark (as well as the remarks above) is involved in forthcoming results, as well.\\

To prove the convergence $Du_t\to Du$ in $L^p$ when $p>n$ we first show that this convergence holds pointwise a.e. This part follows basically from the next theorem.

\begin{theorem}\label{classical}
Suppose that $u$ is continuous and bounded function and differentiable at $x\in\rn\,$. Then $D_{\gamma}u_t(x)\to Du(x)\cdot\gamma$ as $t\to 0$ for any direction $\gamma$. 
\end{theorem}
\textit{Proof.} Denote that $Du(x)=:D$. By the differentiability of $u$ at $x$,  
there exists function $\varepsilon_x$ such that $\varepsilon_x(h)\to 0$ as $|h|\to 0$ and 
\begin{equation}\label{muoto2}
u(x+ta)=u(x)+t(D\cdot a)+t\varepsilon_x(ta)|a|\,.
\end{equation}
Suppose that $a_t\in\R u_t(x)$ so that $D_{\gamma}u_t(x)= -D_{\gamma}L(a_t)$. To complete the proof we have to show that $D_{\gamma}L(a_t)\to -D\cdot \gamma$. 
It follows from (\ref{muoto2}) that for every $t>0$, point $a_t$ minimizes the quantity
\begin{equation}\label{muoto}
D\cdot a+\varepsilon_x(ta)|a|+L(a)\,.
\end{equation}
As it was deduced from (\ref{loppu1}) above, $\{ta_t\}_{t>0}$ is bounded. Combining this with the fact that $a_t$ minimizes (\ref{muoto}) and using again the superlinearity of $L$ one obtains that even $\{a_t\}_{t>0}$ is bounded. 

Next we denote by $A$ the set of points $a'$ which are the minimizers of the above quantity without the error term $\varepsilon_x(ta)|a|$, thus every $a'\in A$ minimizes
\begin{equation}
D\cdot a+L(a).
\end{equation} 
Superlinearity of $L$ implies that $A\not=\emptyset$.
Then it clearly holds (recall that $L$ is $C^1$-function) that $DL(a')=-D$ for every $a'\in A$. 

Suppose then that our claim is not true, thus there exists a sequence $a_{t_k}$ such that $D_{\gamma}L(a_{t_k})\not\to -D\cdot \gamma$. Since the sequence $a_{t_k}$ has to be bounded we may assume that it converges, say to $a_0'$. Then 
\begin{align*}
D\cdot a_0'+L(a_0')&=\lim_{k\to\infty}\big(D\cdot a_{t_k}+\varepsilon_x(t_ka_{t_k})|a_{t_k}|+L(a_{t_k})\big)\\
&\leq \lim_{k\to\infty}\big(D\cdot a+\varepsilon_x(t_ka)|a|+L(a)\big)=D\cdot a+L(a)\,
\end{align*}
for every $a\in\rn$, thus $a'_0\in A$ and $DL(a'_0)=-D$. Therefore 
\[D_{\gamma}L(a_{t_k})\to D_{\gamma}L(a_0')=-D\cdot\gamma\,.\] 
This is the desired contradiction.
\hfill$\Box$.\\

\textit{Remark 1.} 
This implies that if $u$ is differentiable almost everywhere, like if $u\in W^{1,p}(\rn)$, $p>n\,$, then $D_{\gamma}u_t\to Du\cdot\gamma$ almost everywhere and $Du_{t_k}\to Du$ a.e for any sequence $t_k\to 0\,$ (here it is reasonable to formulate the latter claim for sequences, see the remarks above).

\textit{Remark 2.} Above theorem accounts for our assumption $L\in C^1\,$ (or strictly convexity of $H$). Dropping this assumption would yield problems with this theorem. The main reason for this is the fact that we really need the above theorem in the form where anything else about the smoothness of $u$ around point $x$ is not known than the differentiability in that single point $x\,$.\\

%\begin{theorem}
%Suppose that $u$ is continuous, bounded and differentiable at $x\in\rn\,$. Then $D_{\gamma}u_t(x)\to Du(x)\cdot\gamma$ as %$t\to 0$ for any direction $\gamma$. 
%\end{theorem}
%\textit{Proof.} By the differentiability of $u$ at $x$,  
%there exists function $\varepsilon_x$ such that $\varepsilon_x(h)\to 0$ as $|h|\to 0$ and 
%\begin{equation}
%u(y)=u(x)+Du(x)\cdot(y-x)+\varepsilon_x(y-x)|y-x|\,.
%\end{equation}
%Suppose then that $y_t$ gives the value of the inf-convolution at $x$ for $t>0$, thus $y_t\in\R u_t(x)$. It follows from %above that for every $t>0$, point $y_t$ minimizes the quantity
%\begin{equation}\label{muoto}
%Du(x)\cdot(\frac{y-x}{t})+\varepsilon_x(y-x)\frac{|y-x|}{t}+L(\frac{y-x}{t})\,.
%\end{equation}
%Let us denote that $a_t=\frac{y_t-x}{t}$. Recall that $L$ grows in infinity asymptotically faster than any linear %mapping. On the other hand, it follows from the boundedness of $u$ that $\varepsilon_x$ can not grow asymptotically %faster than $|Du(x)||x|$. Therefore it is easy to check that set $\{a_t\}_{0<t\leq 1}$ has to be bounded.

%Furthermore, denote by $y_t^*$ the point which minimizes the above quantity without the error term, i.e. $y_t^*$ minimizes
%\begin{equation}\label{defin}
%F_t(y):=\,Du(x)\cdot (\frac{y-x}{t})+L(\frac{y-x}{t})\,.
%\end{equation}
%Again, let us denote $a_t^*=\frac{y^*_t-x}{t}\,$. Then observe that $y\to \frac{y-x}{t}$ is bijective mapping implying %that every $a^*_t$ minimizes the quantity  
%\begin{equation}
%Du(x)\cdot a\,\,+\,L(a)\,.
%\end{equation}
%Now the strict convexity of $L$ implies that actually minimum point of this quantity is unique, denoted by $a_0$, and %$a^*_t=a_0$ for every $t>0$. 

%Correspondingly, it holds for points $a_t$ that $a_t$ minimizes the quantity 
%\begin{equation}\label{includevirhe}
%Du(x)\cdot a+\varepsilon_x(ta)|a|+L(a)\,. 
%\end{equation} 

%Let us then make use of these observations.
%Firstly, we know that $D_iF_t(y_t^*)=0$ since $y_t^*$ minimizes $F_t$. Using the definition of $F_t$ this implies that 
%\begin{equation}
%-DL(\frac{y_t^*-x}{t})=-DL(a_0)=Du(x)
%\end{equation}
%for every $t>0$.
%Above we showed that $Du_t(x)=-DL(\frac{y_t-x}{t})=-DL(a_t)$ (for a.e. $x$) if $y_t\in\R u_t(x)$. 
%Therefore, to prove the claim it is sufficient to show that 
%\begin{equation}
%|DL(a_t)-DL(a_0)|\to 0\,\text{ as }t\to 0\,.
%\end{equation}
%Since $L$ is $C^1$-function we can verify this by showing that $|a_0-a_t|\to 0\,$ as $t\to 0\,$. 
%Suppose, on the countrary that this does not hold. Then there exists a sequence $\{a_{t_i}\}$ which does not converge to %$a_0$ as $t_i\to 0$. By extracting a subsequence(and using compactness), if needed, we may assume that this subsequence %converges to $a'\in\rn$. Now using the continuity of the above expressions and the uniqueness of the minimum point $a_0$ %we get that
%\begin{align*}
%&\lim_{i\to\infty}\,\big(\,Du(x)\cdot a_0+\varepsilon_x(t_ia_0)|a_0|+L(a_0)\,\big)\\
%=&Du(x)\cdot a_0+L(a_0)\,<\, Du(x)\cdot a'+L(a')\\
%=&\lim_{i\to\infty}\,\big(\,Du(x)\cdot a_{t_i}+\varepsilon_x(t_ia_{t_i})|a_{t_i}|+L(a_{t_i}\,\big).
%\end{align*}
%But this clearly contradicts with the assumption that $a_{t_i}$ are minimizers for the quantity (\ref{includevirhe}). 
%This completes the proof.\\
The remaining ingredient of our main result relies to the following rather standard inequality (following e.g. from Poincare inequality, see \cite{A}).
\begin{lemma}\label{perus2}
Suppose that continuous function $u:\rn\to\re$ has locally integrable weak partial derivatives and $p>n\,$. Then 
\[|u(x)-u(y)|\leq\,C(p,n)|x-y|\,\bigg(\avr{B(x,2|x-y|)}|Du|^p\bigg)^{\frac{1}{p}}\]
for every $x,y\in\rn$ such that $B(x,2|x-y|)\subset\Omega$.
\end{lemma}

Using the above inequality we get the desired $p$-integrable majorant for functions $Du_t\,$ when $|DL(x)|\lesssim \frac{L(x)}{|x|}\,$. Below $M$ denotes the classical centered Hardy-Littlewood maximal function.
\begin{theorem}\label{bound}
If $u$ is bounded and continuous and $L$ satisfies $|DL(x)|\leq C\frac{L(x)}{|x|}\,$, then 
\begin{equation}
\sup_{\gamma}|D_{\gamma}u_t(x)|\leq C'\big(M(|Du|^p)(x)\big)^{\frac{1}{p}}\,\text{ for every }t\in]0,1]\,\text{ and }x\in\rn\,.
\end{equation}
\end{theorem}
\textit{Proof.} 
Observe first that by Lemma (\ref{directional}) 
\[\sup_{\gamma}|D_{\gamma}u_t(x)|=\sup_{\gamma}|D_{\gamma}L(a_{\gamma})|=|D_{\gamma}L(a)|\,,\]
for some $a\in\R u_t(x)\,$ (here we also used the fact that $\R u_t(x)$ is compact and $L$ is $C^1$-function). 
Then, if $a=0$, the claim is trivially true since $DL(0)=0\,$.
Suppose then that $a\neq 0$  whence it holds that 
\begin{align*}
u(x+ta)+ tL(a)\leq u(x)\,\,
\Longrightarrow\,\,& \frac{L(a)}{|a|}\leq\frac{|u(x+ta)-u(x)|}{t|a|}\,.
\end{align*}
Then we get by Lemma \ref{perus2} and assumption $|DL(x)|\leq C\frac{L(x)}{|x|}\,$ that
\begin{align*}
|DL(a)|\leq C\frac{L(a)}{|a|} \leq 
&\,CC(p,n)\bigg(\avr{B(x,2|x-y|)}|Du|^p\bigg)^{\frac{1}{p}}\leq C'\big(M(|Du|^p)(x)\big)^{\frac{1}{p}}\,
\end{align*}
for every $x$.
\hfill$\Box$\\

Then we are ready to prove Theorem \ref{lptulokset}.

%\begin{theorem}
%If $p>n$ and $L$ satisfies $|DL(x)|\leq C\frac{L(x)}{|x|}\,$, then 
%\begin{equation}
%\normi{Du_t}_p\leq C'\normi{Du}_p\,.
%\end{equation}
%\end{theorem}
%\textit{Proof.}
%Let us choose $c=\frac{1+\frac{p}{n}}{2}>1\,$ whence it holds that $\frac{p}{c}>n$. Then combining the boundedness of the %maximal operator in $L^c(\rn)$ and Theorem \ref{perus3} we get that
%\begin{align*}
%\normi{Du_t}_p &\leq 2C(n,p)\normi{\big(M(|Du|^{\frac{p}{c}})\big)^{\frac{c}{p}}}_p
%\,=\,2C(n,p)\big(\normi{M(|Du|^{\frac{p}{c}})}_c\big)^{\frac{c}{p}}\\
%&\leq C'(n,p)\big(\normi{|Du|^{\frac{p}{c}}}_c\big)^{\frac{c}{p}}\,=\,C'(n,p)\normi{|Du|}_p\,.
%\end{align*}

\subsection*{Proof of Theorem \ref{lptulokset}}
Let $p>n$, $u\in W^{1,p}(\rn)$ bounded, and suppose that $L$ satisfies $|DL(x)|\leq C\frac{L(x)}{|x|}\,$. We have to show that  
\[\normi{Du_t-Du}_p\to 0\,\,\text{ as }t\to 0\,,\]
and, moreover, that $\normi{Du_t}_p\leq C'\normi{Du}_p$ for every $t>0$.

Let us begin with choosing a sequence $t_k>0$ and $t_k\to 0$ as $k\to\infty$.
Recall that in case $p>n$ function $u$ is a.e. differentiable. Then it follows by Theorem (\ref{classical}) that $Du_{t_k}\to Du$ almost everywhere. Then the first of the claims above follows by Lebesgue dominated convergence theorem if we only find the desired $p$-integrable bound for the sequence $|Du_{t_k}|$. For that, let us choose $c=\frac{1+\frac{p}{n}}{2}>1$ and apply Theorem (\ref{bound}) for exponent $\frac{p}{c}$ ( where $n<\frac{p}{c}<p$ ) to obtain that   
\[|Du_{t_k}|\leq CM(|Du|^{\frac{p}{c}})\big)^{\frac{c}{p}}\,\,\text{ a.e}.\]
Furthermore, using the boundedness of the maximal operator in $L^c(\rn)$ we get  
\begin{align*}
\normi{\big(M(|Du|^{\frac{p}{c}})\big)^{\frac{c}{p}}}_p
\,=\big(\normi{M(|Du|^{\frac{p}{c}})}_c\big)^{\frac{c}{p}}
\leq C'\big(\normi{|Du|^{\frac{p}{c}}}_c\big)^{\frac{c}{p}}\,=\,C'\normi{|Du|}_p\,.
\end{align*}
These formulas also immediately imply the latter claim.
\hfill$\Box\,$\\

We end up this section with the proof of the proposition \ref{pakko}. When combined with Theorem \ref{lptulokset}, it immediately implies Corollary \ref{seuraus1}. 
\begin{proposition}\label{pakko}
If $L$ is a Legendre transform of $H$ (with aforementioned properties) then (\ref{aina}) follows if there exists $C>1$ and $C'>1$ so that
\begin{equation}\label{paha2}
\frac{H(2x)}{H(x)}>\,2C\text{ for every }x\in\rn\,\text{ and }
\end{equation} 
\begin{equation}\label{paha3}
\sup_{\partial B(0,r)}\,H\,< C'\inf_{\partial B(0,r)}\,H\,.
\end{equation}
for every $r>0$.
\end{proposition}

\textit{Proof.}
Suppose that $H$ satisfies the requirements of the proposition. It is easy to check that for every $q\in\rn$ there is unique $p_q$ such that 
\begin{equation}
L(q)=\sup_{p\in\rn} p\cdot q -H(p)\,=\,p_q\cdot q-H(p_q)\,\,\text{ and, moreover, } DL(q)=p_q\,.
\end{equation}
Thus, the claim of the proposition is equivalent with 
\begin{equation}\label{hassu1}
|p_q||q|\leq C''(p_q\cdot q-H(p_q))\,,
\end{equation}
for some constant $C''<\infty$.
Let then $p'_q:=\frac{(p_q\cdot q)q}{|q|^2}$ denote the projection of $p_q$ on a line $\{tq:t\in\re\}$. Because $p_q\cdot q=p_q'\cdot q$ , it follows that $H(p_q)\leq H(p_q')$. Using the convexity of $H$ and assumption (\ref{paha3}) we then conclude that $|p_q|\leq C'|p_q'|\,$. This in turn implies that 
\begin{equation}\label{hassu2}
|p_q||q|\leq C'(p_q\cdot q)\,.
\end{equation}
Furthermore, using (\ref{paha2}) we get that 
\begin{align*}
p_q\cdot q-H(p_q)\geq& \,\frac{p_q}{2}\cdot q-H(\frac{p_q}{2})\geq \frac{p_q}{2}\cdot q-\frac{1}{2C}H(p_q)\\
=&\,\frac{1}{2C}(p_q\cdot q -H(p_q))+\frac{C-1}{2C}(p_q\cdot q)\,.
\end{align*}
Finally, we conclude with 
\begin{equation}\label{hassu3}
p_q\cdot q\leq \frac{2C-1}{C-1}(p_q\cdot q-H(p_q))\,. 
\end{equation}
This completes the proof, since (\ref{hassu1}) follows from (\ref{hassu2}) and (\ref{hassu3}).
\hfill$\Box$\\

This proposition shows that assumptions (\ref{paha2}) and (\ref{paha3}) for $H$ guarantee that Legendre transformation $L$ of $H$ satisfies (\ref{aina}) and Theorem \ref{lptulokset} is usable. 
    
\section{Counterexamples}\label{osa3}
In this section we consider the sharpness of the results we proved in the previous section. This section is splitted in three subsections. The first one deals with assumption $p>n$ while two latter subsections are devoted to the assumption $DL(x)\lesssim \frac{L(x)}{|x|}\,$. 
\subsection*{The case $p\leq n$}
The following example deals with case $p\leq n$ and the most important special case $L(q)=\frac{1}{2}|q|^2=H(q)$. If we only want to show that theorem (\ref{lptulokset}) does not hold in this case, thus the convergence or the boundedness of the gradients $Du_t$ in Sobolev-norm does not hold, we may do this much more simple than the following construction by using suitable radial functions. However, in the following example it is shown that in this case even pointwise convergence may totally fail. 

For this example we first recall that in the case $L(q)=\frac{1}{2}|q^2|$ it follows directly from Corollary \ref{peruskaava} that for a.e. $x\in\rn$ 
\begin{equation}\label{kaavakaava}
Du_t(x)=\frac{x-y^*}{t}\,\,
\end{equation}
for every $y^*$ such that the infimum in (\ref{hopf}) is attained at $y^*$. 

In the core of the construction below is (of course) the fact that functions $u\in W^{1,n}(\rn)$, $n\geq 2$, may be even discontinuous , i.e. they do not need to have continuous representatives. Therefore it is easy to construct a continuous function $u\in W^{1,n}(\rn)$ for which 
\[\liminf_{k\to\infty}\bigg(\sup\big{\{}\frac{u(x)-u(y)}{|x-y|}\,:\,y\in B(x,2^{-k})\setminus B(x,2^{-k-1})\big{\}}\bigg)=\,\infty\,\] 
even for almost every $x\,$. On the other hand, the careful analysis of the definition of $u_t$ suggests that $\liminf_{t\to 0}|Du_t(x)|$ should be typically comparable to this quantity. In the following example this intuition
is verified.

\subsubsection*{Proof of Theorem \ref{ekavasta}}
Let $n\geq 2$ and $L(q)=\frac{1}{2}|q|^2$. We have to find continuous and compactly supported function $u\in W^{1,n}(\rn)$ such that $|Du_t|\to \infty$ in a set of positive measure when $t\to 0$.
Let us begin with defining grid sets $A_k$ in $[0,1]^n$ by 
\begin{equation}
A_k=\bigg(\{\frac{i}{4^k}: 0\leq i\leq 4^k\,,\,i\in\mathbb{Z}\}\bigg)^n\,.
\end{equation}
Then denote by $N_k$ the number of points in $A_k$. Define also radially increasing functions $g^k$ such that $g^k(0)=-\frac{1}{2^k}$,
$\normi{Dg^k}_n\leq\frac{1}{2^kN_k}\,$ and $g^k$ is supported in $B(0,r_k)$. Moreover, radii $r_k$ are chosen to be so small that $r_k<<\frac{1}{4^k}$ and, especially,  
\[\bigg|\bigcup_{k=1}^{\infty}\bigcup_{a\in A_k}B(a,2r_k)\bigg|=:|B|<\frac{1}{2}.\] 
Finally, define 
\begin{equation}
u(x)=\inf_{k\in\na}\bigg(\sum_{a\in A_k}g^k(x-a)\bigg)=:\inf_{k\in\na}f^k(x)\,.
\end{equation}
Observe that the distance between two points in $A_k$ is always greater than $r_k$ which implies that for every $x$ and $k$ there exists at most one $a\in A_k$ such that $g^k(x-a)\neq 0$. Moreover, $u$ is continuous function since the functions $f^k$ are continuous and $|f^k|\leq 2^{-k}$ for every $k$. Furthermore, it is easy to check that  
\begin{align*}
\normi{Du}_{n}\leq \sum_{k=1}^{\infty}N_k\normi{Dg^k}_n\leq \sum_{k=1}^{\infty}N_k\frac{1}{2^kN_k}\,=1\,.
\end{align*}
This implies that $u\in W^{1,n}(\rn)$.

In what follows we prove that $Du_t\to\infty$ a.e in $B^c$.
For the proof, observe first that
\begin{align*}
u_t(x)&=\inf_{y\in\rn}\,\big(\,u(y)+\frac{|x-y|^2}{2t}\,\big)=
\inf_{y\in\rn}\big(\inf_{k\in\na}\big(f^k(y)+\frac{|x-y|^2}{2t}\big)\,\big)\\
&=\inf_{k\in\na}\big(\inf_{y\in\rn}\big(f^k(y)+\frac{|x-y|^2}{2t}\big)\,\big)\,=\,\inf_{k\in\na}f^k_t(x)\,
\end{align*}
for every $x\in\rn$, $t>0$.

Suppose then that $t\in[4^{-j},4^{-j+1}]$ and $k>\frac{3}{4}j=:k_0$. Then there exists point $y^*\in A_{k_0}$ so that $|x-y^*|\leq C_n4^{-k_0}$ whence 
\[\frac{|x-y^*|^2}{2t}\leq C'_n\frac{4^{-2k_0}}{4^{-j}}< C'_n\frac{1}{2^j}.\]
This implies that
\begin{align*}
f^k_t(x)&\geq -\frac{1}{2^k}\geq -\frac{1}{2^{k_0}}+\frac{1}{2^{k_0+1}}=-\frac{1}{2^{k_0}}+\frac{2^{\frac{j}{4}-1}}{2^j}\\ &\geq -\frac{1}{2^{k_0}}+C'_n\frac{1}{2^j}
\geq f^{k_0}(y^*)+\frac{|x-y^*|^2}{2t}\geq f_t^{k_0}(x).
\end{align*}

We conclude that if $t\in[4^{-j},4^{-j+1}]$ then there exists integer $k_0$ such that 
$u_t(x)=f^{k_0}_t(x)$ and $k_0\leq \frac{3}{4}j\,$. Especially, if $u_t(x)$ is achieved in point $y^*$ then it follows from above that $f^k(y^*)\neq 0$ for some $k\leq\frac{3}{4}j$, thus
\begin{equation}\label{helppo}
y^*\in\bigcup_{k=1}^{\frac{3}{4}j}\bigg(\bigcup_{a\in A_k}B(a,r_k)\bigg)\,.
\end{equation}
Suppose then that $u_t(x)$ is achieved in $y^*$ such that 
$|x-y^*|\leq 4^{-\frac{5}{6}j}$ whence it holds that 
  
\begin{align*}
x&\in\bigcup_{k=1}^{\frac{3}{4}j}\bigg(\bigcup_{a\in A_k}B(a,r_k+4^{-\frac{5}{6}j})\bigg)\\
&\subset\bigg(\bigcup_{k=1}^{\infty}\bigcup_{a\in A_k}B(a,2r_k)\bigg)\,\bigcup\,
\bigg(\bigcup_{k=1}^{\frac{3}{4}j}\bigcup_{a\in A_k}B(a,2(4^{-\frac{5}{6}j}))\bigg)\\
&= B\cup \bigcup_{a\in A_{\frac{3}{4}j}}B(a,2(4^{-\frac{5}{6}j}))=:B\cup B_j\,.
\end{align*}
Observe then that if $x\in B^c\cap B_j^c$ then $|x-y^*|>4^{-\frac{5}{6}j}$ and (recall (\ref{kaavakaava}))  
\[|Du_t(x)|\overset{a.e.}{=}\frac{|x-y^*|}{t}>4^{\frac{1}{6}j}\] 
if $t\in[4^{-j}, 4^{-j+1}]\,$. It follows that
 \[\{x\in B^c: Du_t(x)\overset{t\to 0}{\not\to}\infty\}\subset\bigcap_{m=1}^{\infty}\bigcup_{j=m}^{\infty}B_j\,.\]
Finally, observe that $|B_j|\leq C_n(4^{\frac{3}{4}j}4^{-\frac{5}{6}j})^n\leq C_n 4^{-\frac{1}{12}j}$ and so  $\sum_{j=1}^{\infty}|B_j|<\infty$. This guarantees that $Du_t\to\infty$ a.e. in $B^c$.
\hfill$\Box$\\

\subsection*{Exponential but radial growth}
As it was mentioned in the introduction, perhaps the most natural case when assumption 
$|DL(x)|\lesssim\frac{|L(x)|}{|x|}$ does not hold appears when $L$ behaves in infinity like exponential function. 
At first glance it may look that the more faster $L$ grows in infinity the better the local regularity of $u$ is retained for function $u_t$. In certain sense this is true, especially at the points where $u$ is differentiable. However, in case of Sobolev-functions, even if $p>n$, there might exist singular points for the gradient like when $u(x)=|x|^{\alpha}$, where $\alpha>1-\frac{n}{p}$. It turns out that around these singular points the bad behaviour of $L$ at infinity may kill the uniform local $p$-integrability of the derivatives $Du_t$. More precisely, we prove the following theorem: 

\begin{theorem}
Let $p>n\geq 3$ and suppose that $C^1$-function $L:\rn\to[0,\infty]$ coincides with function $e^{|x|}$ when $|x|\geq 1\,$. In $B(0,1)$ suppose that $L$ is strictly convex with $L(0)=DL(0)=0\,$.     
Suppose that functions $u_t$ are determined by $L$ as in (\ref{hopf}).
Then there exists compactly supported function $u\in W^{1,p}(\rn)$ such that $\normi{Du_t}_{p}\to\infty$ as $t\to 0\,$.
\end{theorem} 
\textit{Proof.}
Let us fix $\frac{1}{2}<\alpha<1$ such that $(\alpha-1)p+n>0\,$. Moreover, define for every $k\in\na$ constants $C_k$ by 
\[C_k=\frac{2^{k(\alpha-1+\frac{n}{p})}}{k^{\frac{3}{2p}}}\,.\]
and functions $u^k$ in $\re$ by 
\begin{equation}\label{definit}
u^k(x)=\begin{cases}
        &C_k|x|^{\alpha}\,\, \text{ if } |x|<2^{-k}\,.\\
	& C_k2^{(-k)\alpha}\,\,\text{ otherwise}\,.
       \end{cases}
\end{equation}
Let us then assume that $t\in[2^{-k},2^{-k+1}]$ and consider the function $u^k_t$,
\begin{equation}\label{definit2}
 u^k_t(x)=\inf_{y\in\re}(u(y)+te^{\frac{|x-y|}{t}})\,.
\end{equation}
We are going to show that actually $u^k_t$ coincides with $te^{\frac{|x|}{t}}$ in $B(0,r)$ such that 
\[r\geq t\log(\frac{t^{\alpha-1}C_k}{2})=:r_0\,.\] 
To prove this, suppose that $|x|=r_0\,$. Then it follows that 
\begin{align*}
|(te^{\frac{\cdot}{t}})'(x)|=e^{\frac{|x|}{t}}= e^{\frac{r_0}{t}}=\frac{t^{\alpha-1}C_k}{2}\leq \frac{(2^{-k})^{\alpha-1}C_k}{2}\,\leq \alpha(2^{-k})^{\alpha-1}C_k\,.
\end{align*}
Since we know that the derivative of $tL\big(\frac{x}{t}\big)$ is increasing (respect to $|x|$), the above estimate implies that in $[-r_0,r_0]$ the supremum of $|(te^{\frac{\cdot}{t}})'(x)|$ is less than the infimum of the absolute value of the derivative of $u^k$ in $[-2^{-k},2^{-k}]$. One can easily see that this guarantees the coincidence of functions $u^k_t$ and $te^{\frac{|x|}{t}}$ in $[-r_0,r_0]$ if we only know that 
\[te^{\frac{r_0}{t}}\leq C_k(2^{-k})^{\alpha}\,.\]  
But this follows by substituting the value of $r_0$ and using the assumption $t\leq 2^{-k+1}\,$. 

Observe then that if functions $u^k$ and correspondingly $u^k_t$ above were defined in $\rn$ (instead of $\re\,$) in the same way as in (\ref{definit}) and (\ref{definit2}), we would end up exactly with the same conclusion, thus the coincidence of $u^k_t$ and $te^{\frac{x}{t}}$ in $B(0,r_0)\,$. Let us then estimate the $p$-norm of the derivative of $u_t$:
\begin{align*}
\int_{B(0,r_0)}|Du^k_t(x)|^p\,dx\,&=\int_{B(0,r_0)}\big(e^{\frac{|x|}{t}}\big)^p\,dx\,\gtrsim 
\int_0^{r_0}z^{n-1}e^{\frac{zp}{t}}\,dz\,\\
&\gtrsim\,r_0^{n-1}\int_0^{r_0}e^{\frac{zp}{t}}\,dz\,\gtrsim r_0^{n-1}\frac{t}{p}e^{\frac{r_0p}{t}}
\\&=t^{n-1}\bigg(\log\bigg(\frac{t^{\alpha-1}C_k}{2}\bigg)\bigg)^{n-1}\frac{t}{p}\big(\frac{t^{\alpha-1}C_k}{2}\big)^p\\
&\approx t^{n+(\alpha-1)p}C_k^p\,\bigg(\log\bigg(\frac{t^{\alpha-1}C_k}{2}\bigg)\bigg)^{n-1}\,.
\end{align*}
Here (as well as below) notation $"\gtrsim"$ and $"\approx"$ mean that corresponding equality/inequality is valid up to a constant which does not depend on $t$ or $k$. 
Furthermore, 
\begin{align*}
\int_{\rn}|Du^k(x)|^p\,dx\,&\approx\int_{B(0,2^{-k})}C_k^p|x|^{(\alpha-1)p}
\approx C_k^p\int_0^{t}z^{n-1}z^{(\alpha-1)p}\,dz\,\\
&\approx t^{n+(\alpha-1)p}C_k^p\,\approx\,\frac{1}{k^{\frac{3}{2}}}. 
\end{align*}

Finally, compute that 
\begin{align*}
\bigg(\log\bigg(\frac{t^{\alpha-1}C_k}{2}\bigg)\bigg)^{n-1}&\geq
\bigg(\log\bigg(\frac{2^{(-k+1)(\alpha-1)}C_k}{2}\bigg)\bigg)^{n-1}\\
&=\bigg(\log\bigg(\frac{2^{(k-1)(1-\alpha)}2^{k(\alpha-1+\frac{n}{p})}\,}{2k^{\frac{3}{2p}}}\bigg)\bigg)^{n-1}\\
&\gtrsim\bigg(\log\bigg(\frac{2^{k\frac{n}{p}}}{k^{\frac{3}{2p}}}\bigg)\bigg)^{n-1}
\,\gtrsim k^{n-1}\,. 
\end{align*}

Combining the above estimates and substituting the value of $C_k$ we get that 
\begin{align*}
\int_{B(0,r_0)}|Du^k_t(x)|^p\,dx\,
%\gtrsim k^{n-1}\int_{\rn}|Du^k(x)|^p\,dx\,\\
\gtrsim k^{n-1}C_k^pt^{n+(\alpha-1)p}\approx k^{n-1}\frac{1}{k^{\frac{3}{2}}}\to\infty \text{ when } k\to\infty\,.
\end{align*}
Above in the last convergence we need the assumption $n\geq 3$ (this assumption is commented below). 

Summing up, we constructed a sequence of radial Sobolev-functions $u^k$ so that $(u^k)\to 0$ in $W^{1,p}(\rn)$, functions $u^k$ are constant outside the ball $B(0,2^{-k})$ and so that \text{if} $t_k$ is in range $[2^{-k},2^{-k+1}]$, then the $p$-norm of $Du^k_{t_k}$ tends arbitrarily big as $k$ tends to infinity. To verify our claim, thus the existence of single compactly supported Sobolev-function $u$ for which $\normi{Du_t}_p\to\infty$ as $t\to\infty$, we will use functions $u^k$ in the following way: Define a collection of disjoint balls $B_k=B(x_k,r_k)\subset\rn$ so that $r_k=k2^{-k}\,$ and all these balls are contained in $B(0,R)$, $R>0$. Then define function $u$ so that outside the union of these balls $u\equiv 0$ and in every $B(x_k,r_k)$ $u$ coincides with $\tilde{u}^k$, where
\[\tilde{u}^k(x)=u^k(x-x_k)-C_k2^{(-k)\alpha}\,.\]
That is, $\tilde{u}^k$ is given by translation of $u^k$ so that $\tilde{u}^k=0$ in $B(x_k,r_k)\setminus B(x_k,2^{-k})\,$.
Then it is easy to check that if $t\in [2^{-k},2^{-k+1}]$ then $u_t$ coincides in $B(x_k,r_k)$ with $u^k_t$ up to the appropriate translation. To guarantee this fact we defined $r_k=k2^{-k}$ to get enough space around every translated $u^k$. 

 Then what we computed above implies that $\normi{Du_t}_p\to\infty$ as $t\to 0\,$. 
The only thing we still have to check is that $\normi{Du}_p$ is finite. But this follows since 
\begin{equation}\label{last}
\normi{Du}_p\leq \sum_{k=1}^{\infty}\normi{Du^k}_p\leq \sum_{k=1}^{\infty}\frac{1}{k^{\frac{3}{2}}}\,<\infty\,. 
\end{equation}
\hfill $\Box$

The above theorem was proved only in the case $n\geq 3$. How about the cases $n=2$ or $n=1$? The one-dimensional case appears as a special case in many ways. Indeed, we state the following conjecture:
\begin{conjecture}
If $n=1$ and $L$ satisfies the assumptions given in the introduction, then $\normi{u'_t}_p\leq C\normi{u'}_p\,$ for any $1\leq p\leq\infty$.
\end{conjecture}
As we proved, in multidimensional case the above theorem may fail if $L$ behaves in infinity like exponential function. In the core of the argument was the fact that exponential function $L$ may 'spread' the singularity of the derivative in such a way that the measure of the set for which the bad behaviour of the gradient is copied becomes enough big to destroy the uniform $p$-integrability.
This phenomenon can not happen in one-dimensional case. However, we predict that proving the above conjecture requires a subtle argument since it seems that simple maximal-function type pointwise inequalities do not exist.

In case $n=2$ it is easy to check that above proof implies that there exists compactly supported $u\in W^{1,p}(\re^2)$ so that 
\[\limsup_{t\to 0}\normi{Du_t}_{p}\to\infty\,.\] 
However, the stronger divergence in this case seems to require 'less rough' constructions and estimates than it was used in the proof of the previous theorem. The difficulty lies in the fact that to make $\normi{Du_t}$ big it seems technically challenging to avoid of using a sequence of 'singularity functions' $u^k$ with pairwise disjoint supports as in above construction ('every scale needs its own function'). Eventually, this kind of construction requires that $\sum \normi{Du^k}_p$ is finite, say typically $\normi{Du^k}_p\lesssim\frac{1}{k}$. 
But this leads to the situation where the increment in the Sobolev-norm of the gradient, caused basically by the 'measure factor' is comparable to $k^{n-1}$ which in this case equals to $k$. In this case factors $k$ and $\frac{1}{k}$ cancel and one does not get the desired divergence. Summing up, we state the following problem:

\begin{question}
Let $1<p<\infty$ and define functions $u_t$ using proper function $L$ such that $L(x)=e^{|x|}$ when $r\geq r_0$. Does there exist compactly supported function $u\in W^{1,p}(\re^2)$ such that $\normi{Du_t}_p\to\infty$ as $t\to\infty\,$? Furthermore, one can ask if this $u$ can be even radial (this question can be posed in the case $n\geq 3$, as well).
\end{question}

\subsection*{Quasi-radiality}
We end up with a simple example relating to the necessity of the assumption (\ref{paha3a}) for our results. Suppose that $s>s'>1$, and define strictly convex function $L$ in the plane by $L(x,y)=|x|^s+|y|^{s'}\,$. Elementary calculations show that now $L$ is Legendre transform of Hamiltonian $H$, where
\begin{align*}
H(x,y)&=\bigg(\frac{s-1}{s}\bigg)\bigg(\frac{1}{s}\bigg)^{\frac{1}{s-1}}|x|^{\frac{s}{s-1}}\,+\,\bigg(\frac{s'-1}{s'}\bigg)\bigg(\frac{1}{s'}\bigg)^{\frac{1}{s'-1}}|y|^{\frac{s'}{s'-1}}\,\\
&=:C(s)|x|^{\frac{s}{s-1}}\,+\,C(s')|y|^{\frac{s'}{s'-1}}\,.
\end{align*}  
We consider this as the most natural example of the failure of (\ref{paha3a}) and in what follows we will show that if $L$ is as above then for any $2<p<\infty$ there exists compactly supported and radial function $u\in W^{1,p}(\re^2)$ such that $\normi{Du_t}_p\to \infty$ as $t\to 0$. Unless our example is in $\re^2$ it is obvious that the same phenomenon can take place also in higher dimensions.
 
Let $0<\alpha<1$ and suppose that function $u$ is compactly supported such that 
\begin{equation}
u(x,y)=|(x,y)|^{\alpha}\, \text{ if }x\in B(0,1)
\end{equation}
and $u$ is smooth outside the origin. Then, if $p>2$, $u\in W^{1,p}(\rn)$ exactly when $\alpha>\frac{p-2}{p}$. We claim that now
\[\normi{u_t}_p\to\infty\text{ as }t\to 0\,,\] 
if $\alpha$ is chosen such that $\alpha-\frac{p-2}{p}$ is positive but small enough. It turns out that it suffices to consider the set of those $(x,y)$ for which  
%The crucial point is that there exist $c_1>0$ and $c_2>0$, which especially do not depend on $t$, such that 
\begin{equation}\label{coincide}
u_t(x,y)=\frac{|x|^s}{t^{s-1}}+\frac{|y|^{s'}}{t^{s'-1}}\,. 
\end{equation}
In this set the infimum in (\ref{hopf}) is achieved at $y=0$ (or alternatively, $\frac{-x}{t}\in\mathcal{R}f(x)\,$).
We claim that there exist $c_1>0$ and $c_2>0$, which especially do not depend on $t$, such that (\ref{coincide}) holds if $(x,y)\in Q$, where  
\begin{equation}\label{crucial}
Q=\big{\{}(x,y)\in\re^2\,:\,|x|\leq c_1t^{\frac{s-1}{s-\alpha}}\,\text{ and }\,|y|\leq c_2t^{\frac{s'-1}{s'-\alpha}}\,\big{\}}. 
\end{equation}
To show this, observe that if (\ref{coincide}) does not hold, then there exists $a\in\re^2$ such that
\begin{equation}\label{alku}
|a|^{\alpha}+\frac{|a_1-x|^s}{t^{s-1}}+\frac{|a_2-y|^{s'}}{t^{s'-1}}<\frac{|x|^s}{t^{s-1}}+\frac{|y|^{s'}}{t^{s'-1}}\,.
\end{equation}
It is easy to check that we may assume that above $|a_1|\leq |x|$ and $|a_2|\leq |y|$. Then it follows from (\ref{alku}) that
\begin{align*}\label{alku2}
|a_1|^{\alpha}+|a_2|^{\alpha}\lesssim |a|^{\alpha}&<\frac{|x|^s-|x-a_1|^s}{t^{s-1}}+\frac{|y|^{s'}-|y-a_2|^{s'}}{t^{s'-1}}\\
&\lesssim \frac{|a_1||x|^{s-1}}{t^{s-1}}+\frac{|a_2||y|^{s'-1}}{t^{s'-1}}\,,
\end{align*}
implying that either
\begin{align*}
&|a_1|^{1-\alpha}\gtrsim \frac{t^{s-1}}{|x|^{s-1}}\, \text{ or }\,|a_2|^{1-\alpha}\gtrsim \frac{t^{s'-1}}{|y|^{s'-1}}\,.\\
\end{align*}
Finally, using again facts $|a_1|\leq |x|$ and $|a_2|\leq |y|$ we get that either
\begin{align*}
&|x|\gtrsim t^{\frac{s-1}{s-\alpha}}\, \text{ or }\,|y|\gtrsim t^{\frac{s-1}{s-\alpha}}\,.  
\end{align*}
This guarantees that (\ref{coincide}) holds in $Q$ for constants $c_1,c_2>0$, independent on $t$. Then, one can estimate $\normi{Du_t}_{p}$ from below simply by 
\begin{align*}
&\int_{\re^2}|Du_t(x,y)|^p\,d(x,y)\,\geq\int_{Q}|D_xu_t(x,y)|^p\,d(x,y)\,\\
\geq & \int_{0}^{c_1t^{\frac{s-1}{s-\alpha}}}\int_{0}^{c_2t^{\frac{s'-1}{s'-\alpha}}}s^p\frac{|x|^{(s-1)p}}{t^{(s-1)p}}\,dy\,dx
\gtrsim \int_{0}^{c_1t^{\frac{s-1}{s-\alpha}}}t^{\frac{s'-1}{s'-\alpha}}\frac{|x|^{(s-1)p}}{t^{(s-1)p}}\,dx\,\\
\gtrsim & \,t^{\frac{s-1}{s-\alpha}}t^{\frac{s'-1}{s'-\alpha}}\frac{t^{\frac{s-1}{s-\alpha}(s-1)p}}{t^{(s-1)p}}
=t^{\frac{s-1}{s-\alpha}+\frac{s'-1}{s'-\alpha}+\frac{p(s-1)(\alpha-1)}{s-\alpha}}=:t^{\beta}\,.
\end{align*}
By elementary calculations, $\beta<0$ is equivalent with  
\begin{equation}\label{joo}
(s'-1)(s-\alpha)+(s-1)(s'-\alpha)<p(s-1)(1-\alpha)(s'-\alpha)\,.
\end{equation}
We are ready if we only can show that $\beta<0$ for suitable $1>\alpha>\frac{p-2}{p}$. This in turn follows if 
(\ref{joo}) holds with substitution $\alpha=\frac{p-2}{p}$. Then $1-\alpha=\frac{2}{p}$ and (\ref{joo}) is equivalent with
\begin{align*}\label{joo2}
&(s'-1)(s-\alpha)+(s-1)(s'-\alpha)<2(s-1)(s'-\alpha)\,\\
\Longleftrightarrow\,\,&(s'-1)(s-\alpha)<(s-1)(s'-\alpha)\,\\
\Longleftrightarrow\,\,&-\alpha s'-s<-\alpha s-s'\,\\
\Longleftrightarrow\,\,&\alpha =\frac{p-2}{p}<1\,.
\end{align*}
This verifies our claim. 

%It turns out that there exists $c_1$
%\begin{equation}
%u_t(x)\frac{|x|^s}{t^s-1}
%\end{equation}
  




%Let then $r>0$ and consider the simple cone function 
%\begin{equation}\label{viimeinen1}
%u^r(x)=\begin{cases}
%        &-r^{1-\frac{2}{p}}+r^{-\frac{2}{p}}|x| \text{ if } |x|<r\,,\\
%	&0\,\,\text{ otherwise}\,.
%       \end{cases}
%\end{equation}
%and suppose that $t>0$ lies in $[\frac{r}{2},r]$. It turns out that now there exists constants $c_1,c_2>0$ (which, in %particular, do not depend on $r$) such that 
%\begin{equation}
%u^r_t(x,y)=-r^{1-\frac{2}{p}}+tL\big(\frac{x}{t},\frac{y}{t}\big)=-r^{1-\frac{2}{p}}+\frac{|x|^s}{t^{s-1}}+\frac{|y|^{s'}}%{t^{s'-1}}
%\end{equation}
%if 
%\begin{equation}
%|x|\leq c_1r\,\,\text{ and }\,\,|y|\leq c_2r^{1-\frac{2(s-s')}{s'p(s-1)}}=:c_2r^{1-\alpha}.
%\end{equation}
%Integrating $|D_xu^r_t|^p=\frac{1}{t^{(s-1)p}}s^p|x|^{(s-1)p}$ over this region implies that 
%\begin{equation}
%\int_{B(0,r^{1-\alpha})}|Du^r_t(y)|^p\,dy\,\geq c'r^{-\alpha}\,.
%\end{equation}
  
 


\begin{thebibliography}{what}
\bibitem[A]{A} R.A.Adams, Sobolev spaces, Academic Press, New York, 1975.
\bibitem[BDR]{BDR} Stefano Bianchini, Camillo De Lellis, Roger Robyr, \textit{SBV Regularity for Hamilton-Jacobi Equations in $\rn$}, Arch. Rational Mech. Anal. 200 (2011), 1003-1021.
\bibitem[H]{H}N. Hoang, \textit{The regularity of generalized solutions of Hamilton-Jacobi equations},
Nonlinear Analysis, 59 (2004) 745-757
\bibitem[JJ]{JJ} V.Julin, P.Juutinen, \textit{A new proof for the equivalence of weak and viscosity solutions for the $p$-Laplace equation}, Communications in Partial Differential Equations, 37 (2012), no.5, 934-946.
\bibitem[K]{K}S.N. Kruzkov, \textit{Generalized solutions of nonlinear first order equations with several
variables}. II. Math. Sbornik, 1 (1967), 93-116.
\bibitem[PR]{PR} J.P.Penot, R.Ratsimahalo, \textit{On the Yosida approximation of operators}, Proceedings of the Royal Society of Edinburgh, 131 A (2001), 945-966.
\bibitem[St]{St} T.Str�mberg, \textit{The Hopf-Lax formula gives the unique viscosity solution}, Differential Integral Equations 15 (2002), no.1, 47-52.
\bibitem[VHT]{VHT} T.D. Van, N. Hoang, M. Tsuji, \textit{On Hopf's formula for Lipschitz solutions of the
Cauchy problem for Hamilton-Jacobi equations}, Nonlinear Anal. 29 (10) (1997) 1145-
1159.
\bibitem[YTW]{YTW}Y.Zhao, T.Tang, J.Wang, \textit{Regularity and global structure of solutions to Hamilton-Jacobi equations I: Convex Hamiltonians}, Journal of Hyperbolic Differential equations 5 (2008), no.3, 663-680.
\end{thebibliography}



% It turns out that for simple 'overturned and cutted cone function' $u$ the $L^p$-norm of $Du_t$ may be as large as we %want when compared to $\normi{Du}_p$. However, for this it is needed that the shape of the cone is suitable chosen for %given $t>0$. Moreover, we need some space around the cone. This leads us to construct a suitable sequence of cone %functions, centered at different points $x_0,x_1,x_2\dots$ and such that they are sufficiently separated. Then we define %our function $u$ as a sum of these cone functions. Of course we have also to make sure that the sum of the Sobolev-norms %of the cone functions is finite. 

%For the true proof, let us define for every $k\in\na$ function $u^k$ by 
%\begin{equation}\label{cones}
%u^k(x)=\min\{0,C_k(-2^{-k}+|x|)\}\,,
%\end{equation}
%where $C_k=\big(\frac{2^{kn}}{k^{1+\varepsilon}}\big)^{\frac{1}{p}}$. Let then $t\in [2^{-k}, 2^{-k+1}]$. We first prove %that for $t$ in this range it holds that
%\[u^k_t(x)=-C_k2^{-k}+tL(\frac{x}{t})\,,\]  
%when $|x|\in[0,z_0]$, where $z_0=t\log(\frac{C_k}{k^2})\,$. For this, suppose $|x|< t\log(\frac{C_k}{k^2})$ and 
%compute that
%\begin{align*}
%|\nabla(tL(\frac{\cdot}{t}))(x)|&=|\nabla %L(\frac{x}{t})|=2\big(\frac{|x|}{t}\big)e^{\frac{|x|}{t}}+\big(\frac{|x|}{t}\big)^2e^{\frac{|x|}{t}}\\
%&< 2(\log\big(\frac{C_k}{k^2}\big))\frac{C_k}{k^2}+\big(\log\big(\frac{C_k}{k^2}\big)\big)^2\frac{C_k}{k^2}
%\leq C_k\frac{\big(\log\big(\frac{C_k}{k^2}\big)\big)^2}{k^2}\,\leq C_k\,. 
%\end{align*}
%The last inequality above can be obtained by substituting $C_k\leq 2^{k}$ inside the logarithm. 
%From this fact one easily deduces that when $|x|\leq z_0$ then $\R u^k_t(x)=\{0\}$ if it holds that 
%\begin{equation}
%-C_k2^{-k}+tL(\frac{|x|}{t})<0\,
%\end{equation}
%when $|x|<z_0$. But this actually follows by the same calculation as above. Let us then estimate 
%\[\int_{B(0,z_0)}|Du^k_t(x)|\,dx\,=\int_{B(0,z_0)}|\nabla L(\frac{x}{t})|^p\,dx\,.\]
%One can straightforwardly to check that this integral is, up to a uniform constant (not depending on $k$), greater than
%\begin{align*}
%z_0^{n-1}\big(\log\big(\frac{C_k}{k^2}\big)\big)^{2p}\int_{0}^{z_0}e^{\frac{zp}{t}}\,&\gtrsim %z_0^{n-1}\big(\log\big(\frac{C_k}{k^2}\big)\big)^{2p}\frac{t}{p}e^{\frac{z_0p}{t}}\\
%&=t^{n-1}\big(\log\big(\frac{C_k}{k^2}\big)\big)^{n-1}\big(\log\big(\frac{C_k}{k^2}\big)\big)^{2p}\frac{t}{p}\big(\frac{C_%k}{k^2}\big)^p\,\\
%&\gtrsim t^nC_k^p\big(\log\big(\frac{C_k}{k^2}\big)\big)^{n-1+2p}\frac{1}{k^{2p}}\\
%&\gtrsim %2^{-kn}\frac{2^{kn}}{k^{1+\varepsilon}}\big(\log\big(\frac{2^{\frac{kn}{p}}}{k^{\frac{1+\varepsilon}{p}}k^2}\big)\big)^{n-%1+2p}\frac{1}{k^{2p}}\\
%&\gtrsim\frac{k^{n-1+2p}}{k^{2p+1+\varepsilon}}\to\infty\,\text{ as }k\to\infty\,,
%\end{align*} 
%since we assumed that $n\geq 3\,$.
%This can be verified by two facts; firstly if 
%\begin{theorem}
%Let $0<\alpha<1$, $1<p<\infty$ and
%\begin{equation}\label{eksponentti}
%L(x)=|x|^2e^{e^{|x|}}\,\,\text{ and }\,\,u(x)=|x|^{\alpha}\,\,\,0<\alpha<1\,.
%\end{equation}%where $\phi$ is compactly supported smooth function and equals to one in $B(0,1)$.  
%Then $\normi{Du_t}_{p,B(0,1)}\to\infty$ as $t\to 0\,$.
%\end{theorem} 
%\textit{Proof.} Let us define
%\begin{equation}\label{kriittinen}
%r_t=\sup\{r>0:u_t(x)=tL\big(\frac{x}{t})\text{ in }B(0,r)\}.
%\end{equation}
%We will verify the claim by showing that for every $t>0$ it holds that
%\begin{equation}\label{second}
%\int_{B(0,r_t)}|D(u_t)|^p\,=
%\int_{B(0,r_t)}|D\big(tL\big(\frac{x}{t}\big)\big)|^p\,\,=
%\int_{B(0,r_t)}|DL\big(\frac{x}{t}\big)|^pdx\,\to\infty\,, 
%\end{equation}
%as $t\to 0\,$. The difficulty here is that it is not very easy to calculate the exact value of $r_t$. However, it is not %too hard to get the sufficiently good estimate. Remark that it is clear that $r_t>0$ by the singularity of the derivative %of $u$ at $0$. For the same reason, it is also easy to get that $\frac{r_t}{t}\to \infty$ as $t\to 0$. 

%Towards the sufficiently good estimate for $r_t$, let us first 
%recall that $L$ and $u$ are radial and monotonic functions implying that functions $u_t$ are radial and monotonic as %well. This allows us to simplify notations by defining functions $g_t$ on $[0,\infty[$ by
%\begin{equation}\label{third}
%g_t(z)=\inf_{0\leq y< z}\big(y^{\alpha}+f_t(z-y)\big)\,,
%\end{equation}
%where  
%\[f_t(z)=t\big(\frac{z}{t}\big)^2e^{e^{\frac{z}{t}}}.\]
%Now it is easy to check that $u_t(x)=g_t(|x|)$ for every $x\in\rn$.
%Remark that definition of $r_t$ can be now written also in form
%\[r_t=\sup\{r>0:\,g_t(z)=f_t(z)\text{ on }[0,r]\,\}.\]  

%Denote again by $\R g_t(z)$ the set of those $y$ which give the infimum in (\ref{third}).
%Consider then the critical point $z=r_t$. By the definition of $r_t$ it follows that $0\in\R g_t(r_t)$. Moreover, it %follows that there exists $y_0>0$ so that 
%$y_0\in\R g_t(x)$ (to prove this one has only use the definition of $r_t$ and the singularity of $(y^{\alpha})'$ at $0$). % The crucial ingredient in our estimate will be the following inequality 
%\begin{equation}
%f'_t(r_t)\geq u'(y_0)=\alpha y_0^{\alpha-1}\,.
%\end{equation}
%It is easy to check that if this would not be true then $g_t$ can not coincide with $f_t$ on $[0,r_t[$, especially on %small interval $[r_t-\varepsilon,r_t[$. On the other hand it holds that $y_0^{\alpha}<f_t(r_t)$ implying that (recall %$\alpha-1<0$) 
%\begin{equation}\label{yht�l�}
%f'_t(r_t)\geq \alpha %y_0^{\alpha-1}=\alpha(y_0^{\alpha})^{\frac{\alpha-1}{\alpha}}>\alpha(f_t(r_t))^{\frac{\alpha-1}{\alpha}}\,
%\end{equation}
%Then we compute that
%\[f'_t(z)=2\big(\frac{z}{t}\big)e^{e^\frac{z}{t}}+(\frac{z}{t})^2e^{\frac{z}{t}}e^{e^{\frac{z}{t}}}\]
%to get that (\ref{yht�l�}) is equivalent with
%\begin{align*}
%2\big(\frac{r_t}{t}\big)e^{e^\frac{r_t}{t}}+(\frac{r_t}{t})^2e^{\frac{r_t}{t}}e^{e^{\frac{r_t}{t}}}&\geq 
%t^{\frac{\alpha-1}{\alpha}}\big(\frac{r_t}{t}\big)^{\frac{\alpha-1}{\alpha}}\bigg(e^{e^{\frac{r_t}{t}}}\bigg)^{\frac{\alph%a-1}{\alpha}}\\
%\Longleftrightarrow \,\,\,
%e^{\frac{1}{\alpha}e^{\frac{r_t}{t}}}\big(2(\frac{r_t}{t})^{\frac{1}{\alpha}}+e^{\frac{r_t}{t}}
%(\frac{r_t}{t})^{\frac{\alpha+1}{\alpha}}\big)\,&\geq\,t^{\frac{\alpha-1}{\alpha}}
%\end{align*}
%Let then $0<\beta=\frac{1-\alpha}{2}<1-\alpha$ and substitute $r_t<t\log(\log(t^{-\beta}))$ to the lefhandside in above inequality. We get that 
%\begin{equation}
%\frac{1}{t^{\frac{\beta}{\alpha}}}\big(2(\log(\log(t^{-\beta})))^{\frac{1}{\alpha}}+\log(t^{-\beta})
%(\log(\log(t^{-\beta})))^{\frac{\alpha+1}{\alpha}}\big)<\frac{1}{t^{\frac{1-\alpha}{\alpha}}}
%\end{equation}
%when $t$ is small. This quarantees that when $t$ is small enough, then $r_t>t\log(\log(t^{-\beta}))$.
%
%Finally, we compute that 
%\begin{align*}
%\int_0^{r_t}(f'_t(z))^p\,dz\,
%\end{align*}

 


 
%Now we can easily estimate the measure of $B$ by
%\begin{align*}
%|B|&\leq C\sum_{k=1}^{\frac{3}{4}j}4^{kn}(r_k+4^{-\frac{5}{6}j})^n\,\leq\,
%2C\sum_{k=1}^{\frac{3}{4}j}(4^kr_k)^n+2C\sum_{k=1}^{\frac{3}{4}j}(4^k4^{-\frac{5}{6}j})^n\\
%&\leq 2C\delta+C_n4^{-\frac{1}{12}j)\,}.
%\end{align*}

%Now it is an easy exercise to show that 
%\begin{equation}
%\nabla u_t(x)=\nabla(f^{k(x,t)}_t)(x)
%\end{equation}
%for almost every $x\in [0,1]^n\,$.

%Following is also needed: If $f^k_t(x)<0$ and $x\not\in\cup_{a\in A_k}B(a,4r_k)$, then
%$f^k_t(x)$ is achieved in point $a_0\in A_k$ with minimal distance from $x$ to the set $A_k$. Thus, 
%\[f^k_t(x)=\frac{1}{2^k}+\frac{|x-A_k|^2}{2t}.\]
%It is easy to see that to prove this it suffices to show the same property for the infimal convolution of function $g^k$.
%More precisely, if $g^k_t(x)<0$ and $|x|>4r_k$ then $g^k_t(x)=\frac{1}{2^k}+\frac{|x|^2}{2t}$. But this can be easily 
%shown by the concavity of the function $g_k$. 
%More precisely, if $f^k_t(x)=u(y)+\frac{|x-y|^2}{2t}<0$, it follows that   
%$\frac{|x-y|}{2t}<\frac{u(x)-u(y)}{}$
%Finally, for every $a\in \bigcup_{k=1}^{\infty} A_k$ let us denote by $\delta(a)$ the smallest integer $k_0$ such that
%$a\in A_{k_0}$.

%\subsubsection*{Conclusion} 
%Let $t\in[4^{-j},4^{-j+1}]$ and $x\in A$. Suppose that $u_t(x)=f^{k(x,t)}_t(x)=:f^{k_0}_t(x)<0$ and $f^{k_0}_t(x)$ is achieved at $y$ (i.e. $y\in\R u_t(x)$) so that $|x-y|\leq 4^{-\frac{9}{10}j}$. First, observe that $k_0\leg j$ by [].

%Because $|x-A_{k_0}|>4r_{k_0}$ we get from above that $y\in A_{k_0}$. Then it holds that within the distance %$C_n4^{-k_0}$ from $y$ lies the point $y'\in A_{k_0-1}$
%and
%\begin{align*}
%&-\frac{1}{2^{k_0}}+\frac{|x-y|^2}{2t}<-\frac{1}{2^{k_0-1}}+\frac{|x-y'|^2}{2t}\,\,\,\Rightarrow\\
%\frac{1}{2^{k_0}}&<\frac{|x-y'|^2}{2t}+\frac{|x-y|^2}{2t}\leq C_n\frac{(4^{-\frac{9}{10}j}+4^{-k_0})^2}{4^{-j}}.
%\end{align*}
%Then, if $k_0>\frac{4}{5}j$ it is easy to compute that  

%For a point $a\in\cup_{k=1}^{\infty}A_k$ denote by $\delta(a)$ the smallest integer $m$ for which $a\in A_m$.




 
%At the points of differentiability it is easy to compute that 

%\begin{align*}
%&(A)\text{ if } x\in\R u_t(x) \text{ then } Du_t(x)=0 \text{ and } \R u_t(x)=\{x\}\,.\\
%&(B) \text{ if } y\in\R u_t(x) \text{ and } y\neq x\text{ then } Du_t(x)=\frac{}{} \text{ and } \R u_t(x)=\{x\}\,.
%\end{align*}
%\[(i)\,\,\text{ There exists unique } x^* \text{ such that } u_t(x)=u(x^*)-\frac{|x-x^*|}{2t}\,\,\text{ and }
%\]


%\subsubsection*{KYSYMYKSI�}
%-Mielenkiintoista?\\
%-Mielenkiintoista kun modifioidaan hieman kysymyksi�?\\
%-Jo tunnettua?\\
%-Eiv�th�n n�m� tulisi jollain lailla triviaalisti tunnetuista ominaisuuksista Hamilton-Jacobi-yht�l�n ratkaisuille?\\
%-Vai onko Hamilton-Jacobi yht�l� mielenkiintoinen vain kun $u$ on Lipschitz, kuten Saksman-Juutinen paperissa oli?
%...

\end{document}
1208.3959.extracted_bib
\begin{thebibliography}{what}
\bibitem[A]{A} R.A.Adams, Sobolev spaces, Academic Press, New York, 1975.
\bibitem[BDR]{BDR} Stefano Bianchini, Camillo De Lellis, Roger Robyr, \textit{SBV Regularity for Hamilton-Jacobi Equations in $\rn$}, Arch. Rational Mech. Anal. 200 (2011), 1003-1021.
\bibitem[H]{H}N. Hoang, \textit{The regularity of generalized solutions of Hamilton-Jacobi equations},
Nonlinear Analysis, 59 (2004) 745-757
\bibitem[JJ]{JJ} V.Julin, P.Juutinen, \textit{A new proof for the equivalence of weak and viscosity solutions for the $p$-Laplace equation}, Communications in Partial Differential Equations, 37 (2012), no.5, 934-946.
\bibitem[K]{K}S.N. Kruzkov, \textit{Generalized solutions of nonlinear first order equations with several
variables}. II. Math. Sbornik, 1 (1967), 93-116.
\bibitem[PR]{PR} J.P.Penot, R.Ratsimahalo, \textit{On the Yosida approximation of operators}, Proceedings of the Royal Society of Edinburgh, 131 A (2001), 945-966.
\bibitem[St]{St} T.Str�mberg, \textit{The Hopf-Lax formula gives the unique viscosity solution}, Differential Integral Equations 15 (2002), no.1, 47-52.
\bibitem[VHT]{VHT} T.D. Van, N. Hoang, M. Tsuji, \textit{On Hopf's formula for Lipschitz solutions of the
Cauchy problem for Hamilton-Jacobi equations}, Nonlinear Anal. 29 (10) (1997) 1145-
1159.
\bibitem[YTW]{YTW}Y.Zhao, T.Tang, J.Wang, \textit{Regularity and global structure of solutions to Hamilton-Jacobi equations I: Convex Hamiltonians}, Journal of Hyperbolic Differential equations 5 (2008), no.3, 663-680.
\end{thebibliography}