Let $p,q$ be conjugate exponents with $p\in(1,\infty]$ and consider a measure space $(X,\mathscr M,\mu)$. Suppose that $S:L^p\to L^p$ and $T:L^p\to L^p$ are both bounded, linear operators. Assume that $S|(L^{p}\cap L^q)=T|(L^{p}\cap L^q)$.
Claim: $S=T$.
The proof is straightforward if $p<\infty$, since $\Sigma$, the set of simple functions supported on finite-measure sets, is dense in $L^p$ and $\Sigma\subseteq L^p\cap L^q$. Hence, since $S$ and $T$ agree on a dense subspace and they are both continuous, they must coincide on the whole space by a limiting argument.
Question: Is this still true if $p=\infty?$ What extra assumptions on $(X,\mathscr M,\mu)$ (e.g. $\sigma$-finitude, semi-finitude, etc.) are needed, if any?
In this case, I think that $L^{\infty}\cap L^1$ is not (necessarily) dense in $L^{\infty}$, so I doubt whether the claim remains true. Yet, I cannot come up with a nice counterexample.
What do you think?
Definitely, $L^\infty\cap L^1$ is not dense in $L^\infty$ if $\mu$ is infinite, since the constant functions are not in the closure of $L^\infty\cap L^1$. So by Hahn–Banach, you can find a bounded linear functional $\phi$ that vanishes on $L^\infty\cap L^1$, yet $\phi(1)=1$.
Now let $S$ be $\phi$ times any fixed member of $L^\infty$, and $T=0$ for a counterexample.
The counterexample must rely on the axiom of choice, unfortunately; you can't get a more concrete example than this.