Let $u:\Omega\subset \mathbb{R}^n\rightarrow \mathbb{R}$ a measureble function. Prove that $u\in L^p(\Omega)$ The function $u:\Omega\rightarrow \mathbb{R}$ is such that
$$
\sup\left\{\int_{\Omega}|u(x)|v(x): v(x)\geq 0 \text{ in }\Omega\text{ and }||v||_{L^q(\Omega)}\leq 1 \right\}
$$
with $\frac{1}{p}+\frac{1}{q}=1$.
I thought about using rever Hölder, because not necessarily $p,q\geq 1$... But, I think that this is an error of the exercise and that we are talking about $p,q\geq 1$.
Later, I thought about using the relation of $L^p(\Omega)$ and $L^q(\Omega)$,
$$
L_v:L^p(\Omega)\rightarrow \mathbb{R},\quad L_v(u)=\int_{\Omega}u(x)v(x)dx.
$$
Because this sup is the norm of operator $L_v$. But, I don't know how to use this, because if I takes on
$$
||L_v||=\sup\left\{\int_{\Omega}|u(x)|v(x): v(x)\geq 0 \text{ in }\Omega\text{ and }||v||_{L^q(\Omega)}\leq 1 \right\},
$$
I will assume that $u\in L^p(\Omega)$.
 A: Let $u_{n}(x)=u(x)\chi_{B_{n}(0)\cap\{|u|\leq n\}}(x)$, then $u_{n}\in L^{p}(\Omega)$ because it is compactyly supported bounded function, and that
\begin{align*}
\sup\left\{\int_{\Omega}|u_{n}(x)|v(x)dx:\|v\|_{L^{q}(\Omega)}\leq 1\right\}\leq M,~~~~n=1,2,...
\end{align*}
where $M$ is the supremum.
Consider the operator
\begin{align*}
T_{n}(v)=\int_{\Omega}|u_{n}(x)|v(x)dx,
\end{align*}
then $\|u_{n}\|_{L^{p}(\Omega)}=\|T_{n}\|\leq M$ for all $n=1,2,...$
By Monotone Convergence Theorem, one has
\begin{align*}
\int_{\Omega}|u(x)|^{p}dx&=\lim_{n\rightarrow\infty}\int_{\Omega}|u_{n}(x)|^{p}dx\leq M^{p}<\infty.
\end{align*}
EDIT:
One need no to recourse to Riesz representation theorem, the operator norm and all that.
In fact, for $u\in L^{p}$, by letting $v=\dfrac{u^{p-1}}{\|u\|_{L^{p}}^{p-1}}$, one has $\int|v|^{q}=\|u\|_{L^{p}}^{(1-p)q}\int|u|^{(p-1)q}=\|u\|_{L^{p}}^{-p}\int|u|^{p}=1$, and that
\begin{align*}
\|u\|_{L^{p}}^{p}=\int|u|^{p}=\|u\|_{L^{p}}^{p-1}\int|u||v|,
\end{align*}
and hence
\begin{align*}
\|u\|_{L^{p}}=\sup\left\{\int|uv|:\|v\|_{L^{q}}\leq 1\right\}.
\end{align*}
