Define for $t > 0$ $$[S(t)g](x) = \int_{\mathbb{R}^n} \Phi(x-y,t)g(y) \, dy \quad (x \in \mathbb{R}^n),$$ where $g : \mathbb{R}^n \to \mathbb{R}$ and $\Phi$ is the fundamental solution of the heat equation. Also set $S(0)g=g$.
(a) Prove $\{S(t)\}_{t \ge 0}$ is a contraction semigroup on $L^2(\mathbb{R}^n)$.
(b) Show $\{S(t)\}_{t \ge 0}$ is not a contraction semigroup on $L^\infty(\mathbb{R}^n)$.
From PDE Evans, 2nd edition: Chapter 7, Exercise 14. I am doing part (b). This question is a continuation of my previous question.
For part (b) of the problem, I am looking for $\|[S(t)]g\|_{L^\infty(\mathbb{R}^n)}=\text{ess sup}|S(t)g|$. Would I need to show that $\|[S(t)]g\|_{L^\infty(\mathbb{R}^n)}=+\infty$ so that the inequality $\|[S(t)]g\|_{L^\infty(\mathbb{R}^n)}\le\|g\|_{L^\infty(\mathbb{R}^n)}$ cannot happen? If so, then this would make $$\|[S(t)]g\|_{L^\infty(\mathbb{R}^n)}=\text{ess sup}|[S(t)g]|=\text{ess sup}\left|\int_{\mathbb{R}^n}\Phi(x−y)g(y)dy\right|=+\infty,$$ I think.
$S(t): L^\infty \to L^\infty$ is indeed a contraction mapping, but $ t \mapsto S(t)u$ need not be continuous in $L^\infty$ at $t = 0$. For example, take $n = 1$ and $$ u(x) = \begin{cases} 1 \quad (x > 0) \\ 0 \quad (x \le 0) \end{cases} \, . $$ Then $S(t)u(x) = \Phi(\frac{x}{\sqrt{t}})$ where $\Phi = S(1)u$. It is easy to see that $\Phi$ is continuous and $\Phi(0) = \frac{1}{2}$. Therefore $S(t)u$ is close to $1/2$ near $x = 0$ for all $t > 0$ and $\|S(t)u - u\|_\infty = \frac{1}{2}$ for all $t > 0$.