Q: The linear operator in $\frac{\partial u}{\partial x} = \lambda u$ is not bounded, but the linear operator in $A u = \lambda u$ is bounded, why? Let $A: R \rightarrow R$ is a linear operator, satisfying $A u = \lambda u$, where $\lambda \in R$, is bounded, and has norm $\|A\| = |\lambda|$.
$\frac{\partial }{\partial x}: C^{\infty}([0,1])\rightarrow C^{\infty}([0,1])$ is also a linear operator, satisfying $\frac{\partial u}{\partial x} = \lambda u$, where $\lambda \in R$ is not bounded, why?
My confusion comes from Example 5.13, Example 5.14 in Hunter's applied analysis.

A: To clarify the examples in Hunter, the operators in question are $A:\mathbb{R}\rightarrow \mathbb{R}$ and $D:C^\infty([0,1])\rightarrow C^\infty([0,1])$ given by $$Ax=ax,$$ where $a\in\mathbb{R}$ is fixed and $$Du(x)=u'(x).$$ The former operator is bounded since, for any non-zero real $x$, $|Ax|/|x|=|a|.$ The operator norm is the supremum over such quotients, and since the quotient is constant in $x$, operator norm is just $|a|$.
The latter operator is not bounded. On a minor variation of Hunter's example, take $u_n(x)=e^{nx}$, where $n\in\mathbb{N}$. Then, $$\|Du_n\|/\|u_n\|=n\rightarrow \infty\text{ as }n\rightarrow\infty.$$ The operator norm is the supremum of such quotients over all non-zero smooth $u$, which we see cannot be finite.
