$ T(g)=0$ for all $g\in L^2\cap L^1$ implies $T(g)=0$ for all $g\in L^2\cap L^p$?

Question

With borel measure.

$T:L^2(\mathbb{R}^n)\cap L^p(\mathbb{R}^n)\to L^p(\mathbb{R}^n)$ continuous operator with the norm $|\cdot |_{p}$, $1<p<\infty$.

If $T(g)=0$ for all $g\in L^2\cap L^1$ Then $T(g)=0$ for all $g\in L^2\cap L^p,\ 1<p<\infty$?

I have this: Because $T$ is continuous and $L^2\cap L^1\cap L^p$ is dense in $L^2\cap L^p$ then holds. It is correct?

Answer 1

Simple functions $g=\sum\limits_{k=1}^{n} c_kI_{E_k}$ with $E_k$'s having finite measure form a dense set in $(L^{2}\cap L^{p}, \|.\|_p)$ and these functions are contained in $L^{2}\cap L^{1}$. Hence $T=0$ on $L^{2}\cap L^{p}$.