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 working on part (a) right now. This following definition on page 435 is relevant:
DEFINITIONS. (i) A family $\{S(t)\}_{t \ge 0}$ of bounded linear operators mapping $X$ into $X$ is called a semigroup if $S(0) u = u$, $S(t+s)u=S(t)S(s)u=S(s)S(t)u$ for all $t,s \ge 0, u \in X$, and the mapping $t \mapsto S(t)u$ is continuous from $[0,\infty)$ into $X$, for all $u \in X$.
(ii) We say $\{S(t)\}_{t \ge 0}$ is a contraction semigroup if in addition $$\|S(t)\| \le 1 \quad (t \ge 0),$$ $\| \quad \|$ here denoting the operator norm. Thus $$\|S(t)u\| \le \|u\| \quad (t \ge 0,u \in X).$$
So I tried to show the very last line of the definition with $[S(t)g](x)$. I did
\begin{align*} \|S(t)g\|_{L^2(\mathbb{R}^n)} &= \left(\int_{\mathbb{R}^n} |[S(t)g](x)|^2 \right)^{1/2} \\ &= \left(\int_{\mathbb{R}^n} \left|\int_{\mathbb{R}^n} \Phi(x-y,t)g(y) \, dy\right|^2 \, dx\right)^{1/2} \\ &\le \left(\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |\Phi(x-y,t)|^2|g(y)|^2 \, dy \, dx\right)^{1/2} \\ &= \left(\int_{\mathbb{R}^n} |\Phi(x-y,t)|^2 \, dx \right)^{1/2} \left( \int_{\mathbb{R}^n} |g(y)|^2 \, dy \, \right)^{1/2} \\ &= \left(\int_{\mathbb{R}^n} |\Phi(x-y,t)|^2 \, dx \right)^{1/2} \|g\|_{L^2(\mathbb{R}^n)} \\ &= (1)^{1/2} \|g\|_{L^2(\mathbb{R}^n)} \\ &= \|g\|_{L^2(\mathbb{R}^n)}. \end{align*} For the second-to-last equality above, I was not sure if I can equate that integral expression to $1$. I did, because the problem says that $\Phi$ is the fundamental solution to the heat equation. And page 46 of the textbook tells that, for each time $t > 0$, $$\int_{\mathbb{R}^n} \Phi(x,t) \, dx = 1,$$ with $$\Phi(x,t) := \begin{cases} \frac 1{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}} & (x \in \mathbb{R}^n,t>0) \\ 0 & (x \in \mathbb{R}^n,t<0). \end{cases}$$
This is a particular case of the Haussdorf-Young inequality. $$\begin{align} |(\Phi\ast g)(x)|&\le\int_{\mathbb{R}^n}\Phi(x-y)\,|g(y)|\,dy\\ &=\int_{\mathbb{R}^n}\Phi(x-y)^{1/2}\,|g(y)|\,\Phi(x-y)^{1/2}\,dy\\ &\le\Bigl(\int_{\mathbb{R}^n}\Phi(x-y)\,|g(y)|^2\,dy\Bigr)^{1/2}\Bigl(\int_{\mathbb{R}^n}\Phi(x-y)\,dy\Bigr)^{1/2}\\ &=\Bigl(\int_{\mathbb{R}^n}\Phi(x-y)\,|g(y)|^2\,dy\Bigr)^{1/2}. \end{align}$$ $$ \int_{\mathbb{R}^n}|(\Phi\ast g)(x)|^2\,dx\le\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\Phi(x-y)\,|g(y)|^2\,dy\,dx=\|g\|_2^2. $$