Let $T$ be a linear and continuous operator $T: L^{2}(\mathbb{R}^{d}) \to L^{2}(\mathbb{R}^{d})$. I wonder to what extent we can conclude if $T$ is strong $(p,p)$, we will obtain that $T'$, the adjoint of $T$, is strong $(p', p')$. For simplicity, we can take strong $(p,p)$ to be restricted to $S(\mathbb{R}^{d})$, that is if $T$ is strong $(p,p)$ iff $\|Tf\|_{L^{p}}\leq \|f\|_{L^{p}}$ for shwarz function $f$.
I think to show $T'$ is strong $(p',p')$ is a duality argument, but I cannot pin down some very specific details, basically have $\sup_{\|f\|_{p}=1, f \text{ simple, compact support}} \langle f, T^{'}g \rangle = \|T^{'}g\|_{p'}$, but this requires $T^{'}g$ to be locally $L^{1}$ integrable, and I guess for the supremum to be finite. If we have $T'(g)$ is shwarz for any function $g$, then of course we have this conclusion by some simple manipulation, but I don't think we can assume that.
Basically, I'm just not very certain what is the exact statement one can make in this circumstance and what needs to be assumed. I saw this in the theory of Calderon Zygmund operator, where from $T$ being strong $(p,p)$ for $1< p< 2$, we have $T'$ is strong $(p',p')$.