Isometric isometry mapping Let $T_g$ be the multiplication operator, then fix $g \in L^\infty(\mathbb{R}^n)$ and let $T_g:L^p(\mathbb{R}^n) \rightarrow L^p(\mathbb{R}^n)$ such that $T_g (f)=g f$. It is fairly simple to show that that $T_g$ is a topological isomorphism whenever there are constants $\alpha,\beta > 0$ such that $\alpha \leq |g(x)| \leq \beta$ almost surely. I am trying to determine for which $g$ will $T_g$ be an isometric isometry, that is a linear isometry that is surjective.
Let $p < \infty$, then to be an isometry we need $\|T_g(f)\|_p^p = \|f\|_p^p$, hence $$\int_{\mathbb{R^n}}|f(x)|^p|g(x)|^p\,dx = \int_{\mathbb{R^n}}|f(x)|^p\,dx.$$ This is satisfied whenever $|g(x)|=1$ almost surely. This also takes care of surjectivity since $1/g$ would be essentially bounded making $T_g$ onto. I believe this takes care of this case. For the case of $p = \infty$ I am running into problems trying to make the equality $\|T_g(f)\|_\infty = \|f\|_\infty$ hold. Any help is appreciated, thanks!
 A: For $p=\infty$ the only such functions $g\in L^\infty$ will be those with $|g(x)|=1$  almost everywhere (i.e. as in the case $p<\infty$). Clearly this condition is sufficient for $T_g$ to be an isometric isometry as in this case we have $|T_g(f)(x)|=|f(x)|$ a.e. Now conversely assume that $\|T_g(f)\|_\infty = \|f\|_\infty$ holds for all $f\in L^\infty$. By choosing  $f=\chi_A$ for measurable sets $A$ we see that $\|g\rvert_A\|_\infty=1$ for all $A$ with positive measure. I claim that this already implies $|g(x)|=1$ a.e. Suppose not, then there is a set $A$ of positive measure such that $|g|<1$ on $A$ (clearly we have $|g|\leq 1$ a.e.). Then there is some $y<1$ and a subset $A'\subseteq A$, still of positive measure, such that $|g|<y$ on $A'$ (consider $A=\bigcup_{n=1}^\infty\{x\in A\mid |g(x)|<1-\frac{1}{n}\}$  to see this). On this subset $A'$ we have $\|g\rvert_{A'}\|_\infty\leq y<1$, a contradiction. Thus $|g(x)|=1$ a.e.
The same argument shows that more generally if $\|f\rvert_A\|_\infty=\|g\rvert_A\|_\infty$ for some $f,g\in L^\infty$ and all $A$, then $f=g$ a.e. A similar statement holds for $p<\infty$.
A: Indeed, it follows $|g| = 1$ almost surely. Here is a more conceptual way to see this.
The spectrum of $T_g$ as an element of the Banach algebra $\Bbb{B}(L^p(\Bbb{R}^n))$ is simply the spectrum of $g$ as an element of the Banach algebra $L^\infty(\Bbb{R}^n)$, i.e. $\sigma(T_g) = \sigma(g)$. Indeed, for $\lambda \in \Bbb{C}$ we have
$$T_{\lambda 1 - g} = \lambda I - T_g \text{ is invertible in $\Bbb{B}(L^p(\Bbb{R}^n))$ } \iff \lambda 1 - g \text{ is invertible in } L^\infty(\Bbb{R}^n).$$
Now, it is easy to show that $\sigma(g)$ is equal to the essential range of $g$, defined as
$$\operatorname{ess. im}(g) = \{y \in \Bbb{R} : \text{ for all $\varepsilon > 0$ we have } \lambda(\{x \in \Bbb{R}^n : |g(x) - a| < \varepsilon\}) > 0\}.$$
It is basically the image of $g$ up to null-sets.
Now, if $T_g$ is assumed to be an isometric isomorphism, then the spectrum of $T_g$ is contained in the unit circle in $\Bbb{C}$. Hence
$$\operatorname{ess. im}(g) = \sigma(g) = \sigma(T_g) \subseteq \mathbb{S}^1.$$
A property of the essential range is that any measurable set $X \subseteq \Bbb{R}^n$ such that $g(X)$ doesn't intersect $\operatorname{ess. im} g$ has $\lambda(X) = 0$. Therefore it follows that the image of $g$ a.e. lies on the unit circle and hence $|g| = 1$ a.e.
