By semigroup theory the following are well known
Suppose that $A:X \longrightarrow X $ is a bounded (or unbounded) operator on some Banach's space $X$, and given the Cauchy problem: $$\frac{d}{dt}u(t)=Au(t), \quad u(0)=u_0$$ the solution is $u(t)=e^{tA}u_0$ for any $u_0\in X$ is unique. The set generated by $\{e^{tA}:t\geq0\}$ is bounded even if $A$ is unbounded.
My question
Consider an example: Given $$\partial_tu=\partial_xu, \quad u(0,x)=u_0(x)$$ and we let $A=\frac{\partial}{\partial x}$ be the differential operator, on $X=L^p$ for $p\geq1$ then $A$ is unbounded.
How do you show that $A$ is unbounded? Thanks in advance
One way is to take a sequence $\{f_k\}$ of functions such that $\|f_k\|$ is some fixed constant but $\|f_k'\|$ becomes large as $n\to\infty$. You may take $f_k(x)=\sqrt ke^{-k^2x^2}$. This is a standard example. Verify that $f_k(x)\in L^2(\mathbb{R})$ but $\|f_k'(x)\|_{L^2}^2\to\infty$ as $n\to\infty$.