An Endpoint Version of Uniform Resolvent Estimate for the Laplacian
Tianyi Ren
Yakun Xi
Cheng Zhang
We prove an endpoint version of the uniform Sobolev inequality in Kenig-Ruiz-Sogge \cite{KRS. Although strong inequality no longer holds for the pairs of exponents that are endpoints in the classical theorem of Kenig-Ruiz-Sogge \cite{KRS, they enjoy restricted weak type inequality. The key ingredient in our proof is an interpolation technique first introduced by Bourgain \cite{Bourgain. Inspired by the proof of the endpoint version, we prove, as a byproduct, that restricted weak type Stein-Tomas Fourier restriction inequality also holds for certain vertices of a pentagon in the interior of which Stein-Tomas holds. Thus by real interpolation, strong Stein-Tomas inequality is valid on part of the boundary of that pentagon.
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\begin{document}
 \title{An Endpoint Version of Uniform Resolvent Estimate for the Laplacian}
 \author{Tianyi Ren, Yakun Xi, Cheng Zhang}
 \maketitle

 \begin{abstract}
 We prove an endpoint version of the uniform Sobolev inequality in Kenig-Ruiz-Sogge \cite{KRS}. Although strong inequality no longer holds for the pairs of exponents that are endpoints in the classical theorem of Kenig-Ruiz-Sogge \cite{KRS}, they enjoy restricted weak type inequality. The key ingredient in our proof is an interpolation technique first introduced by Bourgain \cite{Bourgain}. Inspired by the proof of the endpoint version, we prove, as a byproduct, that restricted weak type Stein-Tomas Fourier restriction inequality also holds for certain vertices of a pentagon in the interior of which Stein-Tomas holds. Thus by real interpolation, strong Stein-Tomas inequality is valid on part of the boundary of that pentagon.
 \end{abstract}
\maketitle
\section{Introduction}
Concerning uniform resolvent estimates, Kenig-Ruiz-Sogge \cite{KRS} proved a classical result, which states that on $\mathbb{R}^n$ where $n\geqslant3$, if the pair of exponents $r$ and $s$ satisfy the conditions

(a)\[\frac{1}{r}-\frac{1}{s}=\frac{2}{n},\]

(b)\[\mathrm{min}\Big\{\Big|\frac{1}{r}-\frac{1}{2}\Big|, \Big|\frac{1}{s}-\frac{1}{2}\Big|\Big\}> \frac{1}{2n},\]
then there exists a constant $C$, depending only on $n$, $r$ and $s$, such that the following inequality holds:\begin{equation}
\|u\|_{L^{s}(\mathbb{R}^n)}\leqslant C\|(\Delta+z)u\|_{L^{r}(\mathbb{R}^n)}, \quad u\in H^{2, r}(\mathbb{R}^n), \quad z\in \mathbb{C}.
\end{equation}
Visualized in the plane whose two axes are $\frac{1}{r}$ and $\frac{1}{s}$, the above permissible exponents constitute the OPEN line segment connecting the point $(\frac{1}{r}, \frac{1}{s})=(\frac{n+1}{2n}, \frac{n-3}{2n})$ and its dual point $(\frac{1}{r}, \frac{1}{s})=(\frac{n+3}{2n}, \frac{n-1}{2n})$. This is illustrated in Figure 1.

\begin{figure}
  \centering
    \includegraphics[height=8.5cm]{1.pdf}
\caption{The open line segment for resolvent estimate}
  \label{1}
\end{figure}

This classical theorem already has important applications to partial differential equations in that it leads to certain unique continuation results. Details are in the second part of \cite{KRS}. Moreover, Kenig-Ruiz-Sogge's theorem is sharp in terms of the exponents - the theorem holds only for exponents lying on the aforementioned open line segment. Here in this paper however, we tackle the endpoints of this open line segment. We prove that although strong estimate like the one in the classical theorem no longer holds for the endpoints, they enjoy restricted weak type estimate. Explicitly, we have the following:

\begin{theorem}
Suppose that the dimension $n\geqslant3$. If $r=\frac{2n}{n+1}, s=\frac{2n}{n-3}$, or $r=\frac{2n}{n+3}, s=\frac{2n}{n-1}$ are one pair of the endpoints, then for any $z\in\mathbb{C}$, the inequality holds: \begin{equation}
\|u\|_{L^{s,\infty}(\mathbb{R}^{n})}\leqslant C\|(\Delta+z)u\|_{L^{r, 1}(\mathbb{R}^n)}
\end{equation} where the constant $C$ is independent of $z\in\mathbb{C}$.
\end{theorem}
In the theorem, the spaces $L^{s, \infty}(\mathbb{R}^{n})$ and $L^{r, 1}(\mathbb{R}^{n})$ mean Lorentz spaces. By real interpolation, this theorem implies Kenig-Ruiz-Sogge's classical result.

The key ingredient in proving our theorem is an interpolation technique first introduced by Bourgain \cite{Bourgain}, and this interpolation technique was brought to our attention by Bak-Seeger\cite{Seeger}. In fact, our proof below is indebted to the inspiration from the work of Bak-Seeger. Combining Bourgain's interpolation technique with a variant of Stein's oscillatory integral theorem is the crucial step giving rise to our restricted weak type estimate. However, Stein's oscillatory integral theorem also results in the important Stein-Tomas Fourier restriction theorem. In Kenig-Ruiz-Sogge \cite{KRS}, this theorem, more precisely its equivalent following from the $TT^*$ argument, is extended to all pairs of exponents in the interior of a pentagon. Applying Bourgain's interpolation technique and Stein's oscillatory integral theorem, we obtain too restricted weak type estimate for certain pairs of exponents that are vertices of the pentagon. Then, real interpolation yields, as a byproduct of our main theorem, strong Stein-Tomas estimate for part of the boundary of that pentagon.

\maketitle
\section{Proof of Theorem 1}
First of all, we need some reductions. It suffices to prove the theorem for one endpoint, say $(\frac{n+1}{2n}, \frac{n-3}{2n})$, because the other follows from duality. Furthermore, noting the gap condition $\frac{1}{r}-\frac{1}{s}=\frac{2}{n}$ on the exponents, we are able to reduce the theorem to the case where $z$ has unit length, $|z|=1$, after a simple rescaling argument. Finally, by continuity, the argument of which we will give at the end of the proof, we may assume that $\mathrm{Im}z\neq0$.

This last reduction enables us to study $(-|\xi|^{2}+z)^{-1}$, whose inverse Fourier transform is the fundamental solution of the operator ``$\Delta+z$'' in our theorem. Therefore, our theorem is a consequence of the following estimate for a multiplier operator:\begin{equation}
\Big\|\Big\{\frac{\hat{u}(\xi)}{-|\xi|^{2}+z}\Big\}^{\widecheck{\ }}\Big\|_{L^{\frac{2n}{n-3}, \infty}(\mathbb{R}^{n})}\leqslant C\|u\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})}.
\end{equation} This in turn, amounts to the inequality for a convolution operator \begin{equation}
\Big\| u(x)\ast \Big\{\frac{1}{-|\xi|^{2}+z}\Big\}^{\widecheck{\ }}(x)\Big\|_{L^{\frac{2n}{n-3}, \infty}(\mathbb{R}^{n})}\leqslant C\|u\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})}.\end{equation}
The proof of the theorem then transforms to the study of the kernel of this convolution operator: $K(x)=\Big\{\frac{1}{-|\xi|^{2}+z}\Big\}^{\widecheck{\ }}(x)$.
The expression for this kernel is actually already in literature, e.g. Gelfand-Shilov \cite{GF}, and by their work, \begin{equation}
K(x)=\Big(\frac{z}{|x|^{2}}\Big)^\frac{n-2}{4}K_\frac{n-2}{2}(\sqrt{z|x|^{2}}),
\end{equation} where \begin{equation} \label{10}
K_{\nu}(w)=\int_{0}^{\infty} e^{-w \mathrm{cosh}t} \mathrm{cosh}(\nu t) \mathrm{d}t
\end{equation} denotes the modified Bessel function. Along with the expression for $K(x)$, we will also need the following facts about the Bessel function, all of which are contained in 
\cite[P. 339]{KRS}.

First, a change of variable $u = e^{t}$ in the expression \eqref{10} for $K_{\nu}(w)$ immediately yields\begin{equation}\label{1}
|K_{\nu}(w)|\leqslant C|w|^{-|\mathrm{Re}(\nu)|},
\end{equation} for $|w|\leqslant1$ and $\mathrm{Re}(w)>0$, where the constant $C$ depends only on $\nu$.
Second, applying the formula (see \cite[P. 19]{Erdelyi}) \begin{equation} \label{9}
\Gamma(\nu+\frac{1}{2})K_{\nu}(w) = \Big(\frac{\pi}{2w}\Big)^{\frac{1}{2}}e^{-w}\int_{0}^{\infty} e^{-t}t^{\nu-\frac{1}{2}}\Big(1+\frac{t}{2w}\Big)^{\nu-\frac{1}{2}} \mathrm{d}t,
\end{equation} which is valid when $\mathrm{Re}\nu \geqslant 0$, we obtain the behavior of the Bessel function for large $|w|$:\begin{equation}\label{2}
|K_{\nu}(w)|\leqslant Ce^{-\mathrm{Re}(w)}|w|^{-\frac{1}{2}}
\end{equation} whenever $|w|\geqslant1$ and $\mathrm{Re}(w)>0$.
Finally, formula \eqref{9} in fact tells us that \begin{equation}\label{3}
K_{\nu}(w)=a_{\nu}(w)w^{-\frac{1}{2}}e^{-w},
\end{equation} for $\mathrm{Re}(w)>0$, where the function $a_{\nu}(w)$ enjoys the decaying property \begin{equation} \label{4}
\Big| \Big( \frac{\partial}{\partial w}\Big)^{\alpha}a_{\nu}(w)\Big| \leqslant C_{\alpha}|w|^{-\alpha}.
\end{equation}

With these preparations, we embark on the task of proving the estimate for the convolution operator $K(x)$. The idea is to treat the part of $K(x)$ inside the unit sphere and the part outside separately, hence we break $K(x)=K_{1}(x)+K_{2}(x)$, where $K_{1}(x)$ is defined to be identical with $K(x)$ in the unit spere $|x|\leqslant1$, and equal $0$ elsewhere.
By estimate \eqref{1}, considering the expression for $K(x)$, we easily obtain that \begin{equation}
|K_{1}(x)|\leqslant C|x|^{-(n-2)}.
\end{equation} Then the desired, indeed strong, inequality \begin{equation} \label{5}
\|u(x)\ast K_{1}(x)\|_{L^{\frac{2n}{n-3}}(\mathbb{R}^{n})}\leqslant C\|u\|_{L^{\frac{2n}{n+1}}(\mathbb{R}^{n})}
\end{equation} follows from Hardy-Littlewood-Sobolev inequality whenever the dimension $n>3$, noticing that the exponents $(\frac{n+1}{2n}, \frac{n-3}{2n})$ are on the line $\frac{1}{r}-\frac{1}{s}=\frac{2}{n}$. If $n=3$ however, we cannot apply Hardy-Littlewood Sobolev inequality, because one exponent $\frac{2n}{n-3}$ is $\infty$ then. Nevertheless, the restricted weak type estimate \begin{equation} \label{12}
\|u(x)\ast K_{1}(x)\|_{L^{\infty}(\mathbb{R}^{3})} \leqslant \|u\|_{L^{\frac{3}{2}, 1}(\mathbb{R}^{3})}\|K_{1}\|_{L^{3, \infty}(\mathbb{R}^{3})} \leqslant C\|u\|_{L^{\frac{3}{2}, 1}(\mathbb{R}^{3})}
\end{equation} still holds, by the Holder's inequality for Lorentz spaces. See for instance, \cite[Theorem 3.6]{O'Neil}.

After that, we turn to our analysis of $K_{2}(x)$, the part of $K(x)$ away from the origin. Applying \eqref{2} yields the estimate\begin{equation}
|K_{2}(x)|\leqslant C|x|^{-\frac{n-1}{2}}e^{-|x|\mathrm{cos}(\frac{1}{2}\mathrm{arg}z)}.
\end{equation}
Because of the exponential term, which may have the decaying property we desire, it seems preferable to separate the case where $\mathrm{arg}z\in [-\frac{\pi}{2}, \frac{\pi}{2}]$, from the case where $\mathrm{arg}z\notin [-\frac{\pi}{2}, \frac{\pi}{2}]$. For the former situation, as just mentioned, the effect of the exponential decay yields the strong estimate\begin{equation} \label{6}
\|u(x)\ast K_{2}(x)\|_{L^{\frac{2n}{n-3}}(\mathbb{R}^{n})}\leqslant C\|u\|_{L^{\frac{2n}{n+1}}(\mathbb{R}^{n})},
\end{equation} which is a result from Young's inequality.

The difficult situation is the latter one, and this is where Bourgain's interpolation technique comes into play. As pointed out in the introduction, this interpolation technique first appeared in \cite{Bourgain} when Bourgain was proving an endpoint bound for the spherical maximal function, and we first noticed it in the proof of an endpoint version of the Stein-Tomas Fourier Restriction Theorem in Bak-Seeger\cite{Seeger}. There is also an elaboration on the abstract theory, developed for fairly general normed vector spaces, in Carbery-Seeger-Wainger-Wright \cite{Carbery}. We are not going into such abstractness and generality, but would rather state the result in our specific setting, that of $L^{p}$ spaces.
\begin{lemma}
Suppose that an operator $T$ between function spaces is the sum of the operators ${T_{j}}:$\[T=\displaystyle\sum_{j=1}^{\infty}T_{j}.\] If for $1 \leqslant p_{1}, p_{2}, q_{1}, q_{2} \leqslant \infty$, there exist $\beta_1,\beta_2 > 0$ and $M_{1}, M_{2} > 0$ such that each $T_{j}$ satisfies
\begin{equation}\|T_j\|_{L^{p_1}\rightarrow L^{q_1}}\le M_12^{-j\beta_1},\end{equation} 
and
\begin{equation}\|T_j\|_{L^{p_2}\rightarrow L^{q_2}}\le M_22^{j\beta_2},\end{equation} 
then we have restricted weak type estimate for the operator $T$ between the intermediate spaces:\begin{equation}
\|Tf\|_{L^{q, \infty}}\leqslant M\|f\|_{L^{p, 1}},
\end{equation} where
\[M=M_{1}^{1-\theta}M_{2}^{\theta},\]

\[\frac{1}{p}=\frac{1-\theta}{p_{1}}+\frac{\theta}{p_{2}},  \frac{1}{q}=\frac{1-\theta}{q_{1}}+\frac{\theta}{q_{2}},\]

\[\theta=\frac{\beta_{1}}{\beta_{1}+\beta_{2}}.\]
\end{lemma}

Having introduced Bourgain's interpolation technique, we return to the proof of the main theorem, dealing with the second situation where $\mathrm{arg}z\notin[-\frac{\pi}{2}, \frac{\pi}{2}]$. By the expression \eqref{3} for the Bessel function, \begin{equation}
K_2(x)=|x|^{-\frac{n-1}{2}}e^{-i|x|\mathrm{sin}(\frac{1}{2}\mathrm{arg}z)}e^{i\frac{n-3}{4}\mathrm{arg}z}e^{-|x|\mathrm{cos}\frac{1}{2}\mathrm{arg}z}a_{\frac{n-2}{2}}(|x|e^{i\frac{1}{2}\mathrm{arg}z}),
\end{equation} where $a_{\nu}(w)$ satisfies the decaying property \eqref{4}.

We dyadically decompose the kernel $K_{2}(x)$. Fix a smooth function $\eta(x)$ that has support in $\{ x:|x|<1 \}$ and is equal to $1$ for $|x|<\frac{1}{2}$. Denote $\delta(x)=\eta(x)-\eta(2x)$. Then let $\beta_{0}(x)=\eta(x)$, and for each $j>1$, let $\beta_{j}(x)=\delta(2^{-j}x)$. It is easy to verify that $\sum_{j=0}^{\infty}\beta_{j}(x)=1$.

For $j\geqslant0$, consider the operator $T_{j}$ given by the kernel $K_{2,j}(x)=\beta_{j}(x)K_{2}(x)$, i.e. $T_{j}u=u\ast K_{2, j}$. To tackle kernels with the appearance like that of $K_{2, j}(x)$, we need invoke the following variant of Stein's oscillatory integral theorem.
\begin{lemma}
Let $n\geqslant3$. Suppose that $1\leqslant p \leqslant 2$, $q=\frac{n+1}{n-1}p'$; in other words, the pair of exponents $(p, q)$ lies on the closed line segment joining $A(\frac{1}{2}, \frac{n-1}{2(n+1)})$ and $(1, 0)$. Then, given a kernel of the form \begin{equation}
L(x)=\delta(x)b(x)e^{i\lambda|x|}|x|^{-\frac{n-1}{2}},
\end{equation} where $\lambda \neq 0$, $\delta(x)$ is a smooth function supported in $\{x \in \mathbb{R}^{n}: \frac{1}{4} \leqslant |x| \leqslant 1\}$, $b(x)\in C^{\infty}(\mathbb{R}^{n})$, and $|((\frac{\partial}{\partial x})^{\alpha}b)(x)|\leqslant C_{\alpha}|x|^{-|\alpha|}$, we have the inequality \begin{equation}
\|L\ast f\|_{L^{q}(\mathbb{R}^{n})}\leqslant C|\lambda|^{-\frac{n}{q}}\|f\|_{L^{p}(\mathbb{R}^{n})},
\end{equation} where the constant $C$ depends only on the function $\delta(x)$ and finitely many of the $C_{\alpha}$ above.
\end{lemma}

This oscillatory integral theorem follows from Stein's oscillatory integral theorem \cite{Stein}. See also \cite[P. 341]{KRS} and \cite[P. 63]{Osc}. It holds for pairs of exponents lying on the closed line segment connecting $(\frac{1}{p}, \frac{1}{q})=(1, 0)$ and $ (\frac{1}{p}, \frac{1}{q})=(\frac{1}{2}, \frac{n-1}{2(n+1)})$. Because of this, we are tempted to interpolate between the intersection of the line segment $\frac{1}{r}-\frac{1}{s}=\frac{2}{n}$ for resolvent estimate with the line segment on which the oscillatory integral theorem holds, and the point $(\frac{2}{n}, 0)$ on the $\frac{1}{r}$ axis. These two points that we want to interpolate between are shown in Figure 2, labeled $P$ and $Q$. It is easy to calculate that $Q$ has coordinates $(\frac{n^{2}+n+2}{2n^{2}}, \frac{n^{2}-3n+2}{2n^{2}})$.

\begin{figure}
  \centering
    \includegraphics[height=8.5cm]{2.pdf}
\caption{The interpolation diagram}
  \label{2}
\end{figure}

There is, however, a small trouble with the three dimensional case. When $n=3$, $P$ actually coincides with our target pair of exponents $(\frac{n+1}{2n}, \frac{n-3}{2n})$, hence interpolation cannot produce the desired result. Fortunately, this is not too much of a trouble, and we are able to remedy it by interpolating between two other points. In the following, we treat the case $n > 3$ first, postponing the exceptional three dimensional case until later.

Here is the interpolation process. At $P$, since $K_{2,j}(x)\leqslant C|x|^{-\frac{n-1}{2}}$ by \eqref{2} and $K_{2,j}(x)$ is supported in $\{x:2^{j-2}\leqslant|x|\leqslant2^{j}\}$, Young's inequality yields \begin{equation}\begin{aligned}
\|K_{2,j}(x)\ast u(x)\|_{L^{\infty}(\mathbb{R}^{n})}&
\leqslant \|u\|_{L^{\frac{n}{2}}(\mathbb{R}^{n})}\|K_{2,j}\|_{L^{\frac{n}{n-2}}(\mathbb{R}^{n})}\\
&\leqslant C2^{-j\frac{n-1}{2}}(2^{jn})^{\frac{n-2}{n}}\|u\|_{L^{\frac{n}{2}}(\mathbb{R}^{n})}\\
&= C 2^{j\frac{n-3}{2}}\|u\|_{L^{\frac{n}{2}}(\mathbb{R}^{n})}.\end{aligned}
\end{equation}
At $Q$, we seek to prove \begin{equation} \label{11}
\|K_{2,j}(x)\ast u(x)\|_{L^{\frac{2n^{2}}{n^{2}-3n+2}}(\mathbb{R}^{n})} \leqslant C2^{-\frac{1}{n}j}\|u\|_{L^{\frac{2n^{2}}{n^{2}+n+2}}(\mathbb{R}^{n})}.
\end{equation} Changing the scale, replacing $x$ with $2^{j}x$, we would be done if we could show \begin{equation}
\|\tilde{K}_{2,j}(x)\ast u(x)\|_{L^{\frac{2n^{2}}{n^{2}-3n+2}}(\mathbb{R}^{n})} \leqslant C2^{-\frac{n^{2}-3n+2}{2n}j}\|u\|_{L^{\frac{2n^{2}}{n^{2}+n+2}}(\mathbb{R}^{n})},
\end{equation} where 
\[\tilde{K}_{2,j}(x) = \delta(x)|x|^{-\frac{n-1}{2}}e^{-i2^{j}|x|\mathrm{sin}(\frac{1}{2}\mathrm{arg}z)}b(2^{j}x),\]
\[b(x) = e^{i\frac{n-3}{4}\mathrm{arg}z}e^{-|x|\mathrm{cos}\frac{1}{2}\mathrm{arg}z}a_{\frac{n-2}{2}}(|x|e^{i\frac{1}{2}\mathrm{arg}z}),\]
and $\delta(x)$ is as mentioned in the dyadic decomposition. The kernel $\tilde{K}_{2,j}$ falls witin the hypotheses of the above oscillatory integral theorem, hence applying it, we obtain the inequality \eqref{11} we were seeking, with the constant $C$ independent of $j$.

The estimates at $P$ and $Q$ for the operator $T_{j}$ enables us to utilize Bourgain's interpolation technique, resulting in the desired restricted weak type estimate for the operator $T$, which is the sum of the $T_{j}$. Specifically, what remains to do is only the following elementary computation.

\[\theta=\frac{\frac{1}{n}}{\frac{1}{n}+\frac{n-3}{2}}=\frac{2}{n^{2}-3n+2};\]

\[\Big(\frac{n^{2}+n+2}{2n^{2}}, \frac{n^{2}-3n+2}{2n^{2}}\Big)\cdot(1-\theta)+\Big(\frac{2}{n}, 0\Big)\cdot\theta=\Big(\frac{n+1}{2n}, \frac{n-3}{2n}\Big).\]
Since the last pair of exponents is exactly the one we are aiming at, Bourgain's interpolation technique yields our target estimate:\begin{equation} \label{7}
\|K_{2}(x)\ast u(x)\|_{L^{\frac{2n}{n-3}, \infty}(\mathbb{R}^{n})}\leqslant C\|u\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})},
\end{equation}where, uniform of $z\in\mathbb{C}$, the constant $C$ depends exclusively on the dimension $n$. \eqref{7} together with \eqref{5}, \eqref{12} and \eqref{6} gives the conclusion of the main theorem whenever the dimension $n>3$.

If the dimension $n=3$, as already observed, we are not able to interpolate between $P$ and $Q$, since $P$ coincides with our target pair of exponents $(\frac{2}{3}, 0)$. To remedy this, we interpolate instead between the two points $(0, 0)$ and $(1, 0)$ on the $\frac{1}{r}$ axis, bearing in mind that the oscillatory integral theorem Lemma 2 holds for the latter pair too. Therefore at $(0, 0)$, Young's inequality gives \begin{equation} \begin{aligned}
\|K_{2,j}(x)\ast u(x)\|_{L^{\infty}(\mathbb{R}^{n})}&
\leqslant \|K_{2,j}\|_{L^{1}(\mathbb{R}^{n})}\|u\|_{L^{\infty}(\mathbb{R}^{n})}\\
&\leqslant C2^{-j}\cdot2^{3j}\|u\|_{L^{\infty}(\mathbb{R}^{n})}\\
&=C2^{2j}\|u\|_{L^{\infty}(\mathbb{R}^{n})},\end{aligned}
\end{equation} while also similar to the procedure for the case $n>3$, the oscillatory integral theorem Lemma 2 along with a change of scale argument shows immediately that at $(1, 0)$,\begin{equation}
\|K_{2,j}(x)\ast u(x)\|_{L^{\infty}(\mathbb{R}^{n})}
\leqslant C2^{-j}\|u\|_{L^{1}(\mathbb{R}^{n})}.
\end{equation} The corresponding interpolation process is now like this:

\[\theta=\frac{1}{1+2}=\frac{1}{3},\]

\[(0, 0)\cdot\theta+(1, 0)\cdot(1-\theta)=(\frac{2}{3}, 0).\]
Again, the last pair of exponents is our target pair.

Recall that at the beginning of our proof, we imposed the assumption that $\mathrm{Im}z\neq0$ and stated that the case $\mathrm{Im}z=0$ results from continuity. Indeed, if $z=a$ is real, then by the conclusion we already have, there is a constant $C$, not relying on $b>0$, such that\begin{equation}
\|u\|_{L^{\frac{2n}{n-3}, \infty}(\mathbb{R}^{n})}\leqslant C\|(\Delta+a+ib)u\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})}.
\end{equation} However, by Minkowski's inequality for the space $L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})$, noting that it is a Banach space and indeed has Minkowski's inequality, it follows that the latter term above is majorized by $C\Big[ \|(\Delta+a)u\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})}+b\|u\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})}\Big]$. Letting $b$ go to $0$ yields the case $\mathrm{Im}z=0$.
That concludes our proof of the main theorem.

\maketitle
\section{Endpoint Version of Stein-Tomas Fourier Restriction Theorem}
The original Stein-Tomas Fourier restriction theorem \cite{Stein}, \cite{Tomas} states that if the dimension $n\geqslant3$, one has the inequality \begin{equation}
\Big\|\int_{S^{n-1}}\hat{f}(\xi)\mathrm{e}^{2\pi i\langle x, \xi\rangle}\,\mathrm{d}\sigma(\xi)\Big\|_{L^{\frac{2n+2}{n-1}}(\mathbb{R}^{n})}\leqslant C\|f\|_{L^{\frac{2n+2}{n+3}}(\mathbb{R}^{n})},
\end{equation} for any $f$ belonging to the Schwarz space. Sogge \cite{Osc} extended this result in showing that the same inequality holds for pairs of exponents $(p, q)$ off the line of duality satisfying $1 \leqslant p<\frac{2n}{n+1}, q=\frac{n+1}{n-1}p'$. They constitute the half open line segment in the $(\frac{1}{p}, \frac{1}{q})$ plane connecting $(1, 0) $ and $(\frac{n+1}{2n}, \frac{(n-1)^{2}}{2n(n+1)})$. Therefore by duality and interpolation, the Stein-Tomas restriction inequality is true for pairs of exponents in the interior of the pentagon shown below. However, the cases of the boundary of this pentagon remain unproved.

Considering the fact that Stein-Tomas theorem, including Sogge's version, also follows from Stein's oscillatory integral theorem \cite{Stein}, and inspired by our application of Bourgain's interpolation technique in the proof of the main theorem, we can tackle the boundary case by showing restricted weak type estimate for the vertices of pentagon. Specifically, we prove the following restricted weak type inequality for the point $A(\frac{n+1}{2n}, \frac{(n-1)^{2}}{2n(n+1)})$ \begin{equation}
\Big\|\int_{S^{n-1}}\hat{f}(\xi)\mathrm{e}^{2\pi i\langle x, \xi\rangle}\,\mathrm{d}\sigma(\xi)\Big\|_{L^{\frac{2n(n+1)}{(n-1)^{2}}, \infty}(\mathbb{R}^{n})}\leqslant C\|f\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})},
\end{equation} using Bourgain's interpolation technique. The two points we wish to take into account are $Q (\frac{n^{2}+n+2}{2n^{2}}, \frac{n^{2}-3n+2}{2n^{2}})$, the one that also appears in the interpolation process in the proof of the main theorem, and $E (\frac{1}{2}, \frac{n-1}{2(n+1)})$, one endpoint of Stein's oscillatory integral theorem. See Figure 3.

\begin{figure}
  \centering
    \includegraphics[height=8.5cm]{3.pdf}
\caption{The interpolation diagram for the restriction estimate}
  \label{3}
\end{figure}

Our endpoint version of the Stein-Tomas Fourier Restriction Theorem follows from an analysis of the convolution operator whose kernel is the Fourier transform of the Lebesgue measure on the unit sphere, like in the classical case. This kernel is well-known to have the expression
\begin{equation}
K(x)=2\pi |x|^{-\frac{n-2}{2}}J_{\frac{n-2}{2}}(2\pi|x|),
\end{equation} where $J_{\nu}(w)$ is the Bessel function, see for instance, \cite[P. 347-348]{HA}. Later, we will need the following fact about $J_{\nu}(w)$ for $\nu = m$ positive, integral or half integral, and $w = r$ real, positive and greater than $1$, which is also widely known: it takes the form \begin{equation} \label{8}
J_{m}(r) = \sum_{\pm} r^{-\frac{1}{2}}e^{\pm ir}a_{\pm}(r),
\end{equation} where the functions $a_{\pm}(r)$, $r > 1$ are smooth and satisfy the decay property
\[\Big|\frac{\mathrm{d}^{k}}{\mathrm{d}r^{k}}a_{\pm}(r)\Big| \leqslant C_{k}r^{-k}.\]
This expression can be found in \cite[P. 338]{HA}; see also \cite[Theorem 1.2.1]{fio}. Our aim is to prove the inequality \begin{equation}
\|K\ast f\|_{L^{\frac{2n}{n-3}, \infty}(\mathbb{R}^{n})}\leqslant C\|f\|_{L^{\frac{2n}{n-1}, 1}(\mathbb{R}^{n})}.
\end{equation}

We dyadically decompose $K(x)$, letting $\beta_{j}(x)$, $j \geqslant 0$ be as in the decomposition before and $K_{j}(x)=\beta_{j}(x) K(x)$. $K_{0}(x)$ is easy to handle, since $K(x)$, being the Fourier transform of a distribution with compact support, is smooth, and then the stronger inequality $\|K_{0}\ast f\|_{L^{\frac{2n}{n-3}}(\mathbb{R}^{n})}\leqslant C\|f\|_{L^{\frac{2n}{n-1}}(\mathbb{R}^{n})}$ follows by Young's inequality.

At $Q$ for each $j > 0$, noting the expression \eqref{8} for the Bessel function, a change of scale argument in which we replace $x$ with $2^{j}x$, together with the oscillatory integral theorem Lemma 2 yields the estimate \begin{equation} \begin{aligned}
\|K_{j}(x)\ast f(x)\|_{L^{\frac{2n^{2}}{n^{2}-3n+2}}(\mathbb{R}^{n})}&
\leqslant C2^{j\frac{n-3}{2}}\cdot2^{-j\frac{n^{2}-3n+2}{2n}}\|f\|_{L^{\frac{2n^{2}}{n^{2}+n+2}}(\mathbb{R}^{n})}\\
&=C2^{-\frac{1}{n}j}\|f\|_{L^{\frac{2n^{2}}{n^{2}+n+2}}(\mathbb{R}^{n})}.\end{aligned}
\end{equation} Following the same lines at $E$ for each $j > 0$ gives \begin{equation} \begin{aligned}
\|K_{j}(x)\ast f(x)\|_{L^{\frac{2(n+1)}{n-1}}(\mathbb{R}^{n})}&
\leqslant C2^{j\frac{n^{2}+1}{2(n+1)}}\cdot2^{-j\frac{n(n-1)}{2(n+1)}}\|f\|_{L^{2}(\mathbb{R}^{n})}\\
&=C2^{j\frac{1}{2}}\|f\|_{L^{2}(\mathbb{R}^{n})}.\end{aligned}
\end{equation} We then interpolate between $Q$ and $E$, doing the following string of computation:

\[\theta=\frac{\frac{1}{n}}{\frac{1}{n}+\frac{1}{2}}=\frac2{n+2},\]

\[\Big(\frac{n^{2}+n+2}{2n^{2}}, \frac{n^{2}-3n+2}{2n^{2}}\Big)\cdot(1-\theta)+\Big(\frac{1}{2}, \frac{n-1}{2(n+1)}\Big)\cdot\theta=\Big(\frac{n+1}{2n}, \frac{(n-1)^{2}}{2n(n+1)}\Big).\]

This last pair of exponents is exactly our goal once again. Therefore by Bourgain's interpolation technique, we obtain the desired restricted weak type estimate for the convolution operator with kernel $K(x)$, hence the restricted weak type Stein-Tomas inequality \begin{equation}
\Big\|\int_{S^{n-1}}\hat{f}(\xi)\mathrm{e}^{2\pi i\langle x, \xi\rangle}\,\mathrm{d}\sigma(\xi)\Big\|_{L^{\frac{2n(n+1)}{(n-1)^{2}}, \infty}(\mathbb{R}^{n})}\leqslant C\|f\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})}.
\end{equation} Moreover, duality gives a restricted weak type inequality for the dual pair of exponents $B (\frac{n^{2}+4n-1}{2n(n+1)}, \frac{2n}{n-1})$. With these, we might then apply real interpolation to assert that strong Stein-Tomas inequality is valid on the open line segment connecting $A$ and $B$. Note that this is part of the boundary of the pentagon in which Stein-Tomas holds.

\begin{theorem}
If the dimension $n\geqslant3$, then the Stein-Tomas inequality\begin{equation}
\Big\|\int_{S^{n-1}} \hat{f}(\xi)\mathrm{e}^{2\pi i \langle x, \xi \rangle}\mathrm{d}\sigma(\xi)\Big\|_{L^{q}(\mathbb{R}^{n})}\leqslant C\|f\|_{L^{p}(\mathbb{R}^{n})}
\end{equation} holds for pairs of exponents $(p, q)$ satisfying $\frac{2n(n+1)}{n^{2}+4n-1} < p < \frac{2n}{n+1}$, $\frac{1}{p}-\frac{1}{q}=\frac{2}{n+1}$.
\end{theorem}

One might also apply Bourgain's interpolation technique between the point $(0, 0)$ and the point $(1, 0)$, to get a restricted weak type estimate for the pair of exponents $C(\frac{n+1}{2n}, 0)$, which is also a vertex of the pentagon in which Stein-Tomas holds. Dyadically decompose $K(x)$, notation the same as above. Still, we do not have to worry about the part $K_{0}(x)$ of $K(x)$ near the origin. Away from the origin, i.e., for $j > 0$, at $(0, 0)$, Young's inequality gives \begin{equation} \begin{aligned}
\|K_{j}(x)\ast f(x)\|_{L^{\infty}(\mathbb{R}^{n})}&\leqslant C\|K_{j}(x)\|_{L^{1}(\mathbb{R}^{n})}\|f\|_{L^{\infty}(\mathbb{R}^{n})}\\
&=2^{-\frac{n-1}{2}j}\cdot 2^{nj}\|f\|_{L^{\infty}(\mathbb{R}^{n})}\\
&=2^{\frac{n+1}{2}j}\|f\|_{L^{\infty}(\mathbb{R}^{n})},\end{aligned}
\end{equation}
while at $(1, 0)$, Young's inequality gives \begin{equation} \begin{aligned}
\|K_{j}(x)\ast f(x)\|_{L^{\infty}(\mathbb{R}^{n})}& \leqslant C\|K_{j}(x)\|_{L^{\infty}(\mathbb{R}^{n})}\|f\|_{L^{1}(\mathbb{R}^{n})}\\
&=C2^{-\frac{n-1}{2}j}\|f\|_{L^{1}(\mathbb{R}^{n})}.
\end{aligned}\end{equation}
With the following interpolation computation,

\[\theta=\frac{\frac{n-1}{2}}{\frac{n-1}{2}+\frac{n+1}{2}}=\frac{n-1}{2n},\]

\[(1, 0)\cdot (1-\theta)+(0, 0)\cdot \theta=(\frac{n+1}{2n}, 0),\]
 we obtain the restricted weak type Stein-Tomas inequality at $C (\frac{n+1}{2n}, 0)$: \begin{equation}
\Big\|\int_{S^{n-1}} \hat{f}(\xi) \mathrm{e}^{2\pi i\langle x, \xi\rangle} \mathrm{d}\sigma(\xi)\Big\|_{L^{\infty}(\mathbb{R}^{n})}\leqslant C\|f\|_{L^{\frac{2n}{n+1}, 1}(\mathbb{R}^{n})}.
\end{equation} Duality then produces the same restricted weak type inequality for the pair of exponents $D (1, \frac{n-1}{2n})$. However, this time, we cannot apply real interpolation to the points $A$ and $C$, nor can we apply it to $B$ and $D$, since they are along a vertical or horizontal line, which violates the hypothesis of the real interpolation theorem.

Nevertheless, we can proceed as in the proof of Theorem 1 to obtain a restricted weak type Stein-Tomas inequality at every point on the line segment joining $A$ and $C$ and its dual line segment joining $B$ and $D$. Explicitly, we interpolate between the intersection $S$ of the line $\frac{1}{r}-\frac{1}{s}=k$ ($\frac{2}{n+1} < k <\frac{n+1}{2n}$) with the line $s=\frac{n+1}{n-1}r'$ on which the oscillatory integral theorem Lemma 2 holds, and the point $R(k, 0)$ on the $\frac{1}{r}$ axis. By computation, we find the coordinates of $S$ to be
\[\Big(\frac{n-1}{2n}+\frac{n+1}{2n}k, \frac{n-1}{2n}(1-k)\Big).\]
See the illustrative picture above for these two points. Since the process is the same as before, we omit the detail of calculation, giving only the results.

\begin{figure}
  \centering
    \includegraphics[height=8.5cm]{4.pdf}
\caption{The interpolation diagram for the restriction estimate}
  \label{4}
\end{figure}

At $R$ for each $j > 0$, Young's inequlity leads to the estimate \begin{equation}
\|K_{j}(x)\ast f(x)\|_{L^{\infty}(\mathbb{R}^{n})} \leqslant C2^{j(n(1-k)-\frac{n-1}{2})}\|f\|_{L^{\frac{1}{k}}(\mathbb{R}^{n})},
\end{equation} while at $S$ for $j > 0$, the change of scale argument in which we replace $x$ with $2^{j}x$, together with Lemma 2, produces \begin{equation}
\|K_{j}(x)\ast f(x)\|_{L^{s}(\mathbb{R}^{n})} \leqslant C2^{j(-\frac{n+1}{2}k+1)}\|f\|_{L^{r}(\mathbb{R}^{n})},
\end{equation} where $r$ and $s$ here denote the exponents correponding to $S$. Finally we verify

\[\theta = \frac{\frac{n+1}{2}k-1}{\frac{n-1}{2}(1-k)},\]

\[\Big(\frac{n-1}{2n}+\frac{n+1}{2n}k, \frac{n-1}{2n}(1-k)\Big)(1-\theta)+(k, 0)\theta = \Big(\frac{n+1}{2n}, 1-k-\frac{n-1}{2n}\Big),\]
which gives us a restricted weak type Stein-Tomas inequality at every point on the line segment $AC$, as we hoped. Duality then produces the same result for every point on the dual line segment $BD$.

Having restricted weak type Stein-Tomas inequality for every point on the closed line segment connecting $A$ and $C$, and its dual connecting $B$ and $D$, one might be tempted to ask whether strong Stein-Tomas holds for at least some part of the two lines. Unfortunately this is not true for any of these points. In fact in \cite{Vilenkin}, radial functions belonging to $L^{\frac{2n}{n+1}}{(\mathbb{R}^{n})}$ are constructed that have infinite Fourier transforms on $S^{n-1}$.

To conclude, we may summarize the results we have concerning Stein-Tomas inequality on the boundary of the pentagon. Strong Stein-Tomas holds for every pair of exponents lying on the open line segment connecting $A$ and $B$, for the half open line segment connecting $C$ and $(1, 0)$, and for the half open line segment connecting $D$ and $(1, 0)$. The latter two follows trivially from Young's inequality. Restricted weak type Stein-Tomas holds at every point on the closed line segment joining $A$ and $C$, and its dual joing $B$ and $D$. Finally, counter-examples rule out the possibly of a strong Stein-Tomas at any of the points on $AC$ and $BD$. 

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