Duality argument to extend operators that commute with translations

Question

For $1 \leq p,q \leq \infty$, let $T$ be a linear operator on the space of Schwarz functions $\mathcal S$ that is bounded from $L^p(\mathbb R^n)$ to $L^q(\mathbb R^n)$ and that $T$ commutes with all translations. I have proven that there is a unique $K \in \mathcal S'$ such that $Tf = K*f$ and now am being asked to prove that $T$ can be extended to a bounded linear operator from $L^{q'}$ to $L^{p'}$. I feel like this should follow from a standard duality argument but I've always had a bit of trouble understanding these and am unable to figure this one out.

Answer 1

Let $g \in L^{q'}$ and $f \in L^p$. Note that by H"older, we have $$\left\vert \int (Tf)g \right\vert \leq \int \vert (Tf)g \vert \leq \Vert Tf \Vert_q \Vert g \Vert_{q'} < \infty,$$ since $T$ is bounded from $L^p$ to $L^q$. So, define $\phi_g(f)$ by $$\phi_g(f) = \int (Tf)g,$$ which we know from the above is a bounded linear functional. Hence, there exists a unique $g' \in L^{p'}$ such that $$\phi_g(f) = \int fg'.$$ Define $T$ on $L^{q'}$ by $Tg = g'$. Then, $\Vert Tg \Vert_{p'} = \Vert g' \Vert_{p'} < \infty$.