Partial Converse of Holder's Theorem Holder's Theorem is the following: Let $E\subset \mathbb{R}$ be a measurable set. Suppose $p\ge 1$ and let $q$ be the Holder conjugate of $p$ - that is, $q=\frac{p}{p-1}.$ If $f\in L^p(E)$ and $g\in L^q(E),$ then $$\int_E \vert fg\vert\le \Vert f\Vert _p\cdot\Vert g \Vert_q$$
I am trying to prove the following partial converse is true. Suppose $g$ is integrable, $p>1$ and $$\int_E \vert fg\vert\le M\Vert f\Vert_p$$ when $f\in L^p(E)$ is bounded, for some $M\ge0$. I am trying to prove that this implies that $g\in L^q(E)$, where $q$ is the conjugate of $p$. I am interested in knowing if my proof is correct or if it has any potential.
Attempt: Assuming this is true, I want to show $\int_E \vert g\vert^q<\infty$. Note that $g^{q-1}\in L^p(E)$ since $\int_E \vert g\vert^{pq-p}=\int_E \vert g \vert ^q<\infty$ (I think this is true since $g$ is integrable - I know it holds when $q$ is a natural number). Take $f=g^{q-1}$ and the hypothesis tells us that $$\int_E \vert g \vert^q\le M\Vert g^{q-1}\Vert_P\\<\infty \,\,(\text{since }g^{q-1}\in L^p)$$so $g$ is integrable.

An alternative proof.
Let $T_g(f):=\int_E f g\,,\, \forall f\in L^p.$
It's clear that $T_g$ is linear.
By the condition $\int_E \vert fg\vert\le M\Vert f\Vert_p$, we see $T_g$ is bounded.
So, $T_g\in (L^p)^*.$
By the Riesz Representation Theorem for the Dual of $L^p$(c.f. p. 160, Real Analysis, 4th Edition, Royden), we know there exists an unique $L^q$ function $\tilde{g}$ s.t. $T_g(f)=\int_E f \tilde{g}\,,\, \forall f\in L^p.$  By the uniqueness, we conclude that $g=\tilde{g}\in L^q.$
$\tag*{$\blacksquare$}$

The following is the theorem I quoted in my answer.

$\tag*{$\blacksquare$}$
