$\forall \phi\in C_o^{\infty}(\Omega)$, $\lvert\tilde{f}(\phi)\rvert\le \lVert \phi\rVert_{L^{p'}}\implies f\in L^p$

Question

Let $\Omega\subset\mathbb{R}^d$ open and $\tilde{f} : L^{p'}(\Omega)\to \mathbb{R}$ defined by $\tilde{f}(\phi) :=\int f\phi$ with the property that $\lvert\tilde{f}(\phi)\rvert\le \lVert \phi\rVert_{p'}$ for all $\phi\in C_0^{\infty}(\Omega)$. Now I want to show that $\tilde{f}\in (L^{p'})^*$ so I think I have to show that $\tilde{f}$ is bounded. I almost have this because $\sup\limits_{\phi_\in C_0^{\infty}(\Omega)}\frac{\lvert\tilde{f}(\phi)\rvert}{\lVert \phi \rVert}\le 1$. Also I know that $C_0^{\infty}(\Omega)$ is dense inside $L^{p'}(\Omega)$. But then I am not sure how to show that $\tilde{f}$ is bounded, i.e. $\sup\limits_{g\in L^{p'}(\Omega)}\frac{\lvert\tilde{f}(g)\rvert}{\lVert g \rVert}\le 1$.

Answer 1

I suppose that $f$ is assumed to be a function in $L^1_{\text{loc}}(\Omega)$, that is, $f$ is measurable, and integrable on any compact subset of $\Omega$.

Incidentally this is one of the biggest spaces of functions of interest in Analysis since it contains every $L^p(\Omega)$. But still, if $f$ lies in $L^1_{\text{loc}}(\Omega)$, then $f\phi$ is integrable for every $\phi$ in $C^\infty_c(\Omega)$ (smooth functions with compact support).

Ok, so we are given that the functional $$\tilde f: \phi\in C^\infty_c(\Omega) \mapsto \int_\Omega f\phi\in \mathbb R$$ is continuous relative to the $p'$-norm on $C^\infty_c(\Omega)$, and we need to prove that $f$ lies in $L^p(\Omega)$.

Since $\tilde f$ is continuous, and $C^\infty_c(\Omega)$ is dense in $L^{p'}(\Omega)$, then $\tilde f$ extends uniquely to a continuous linear functional $F$ on $L^{p'}(\Omega)$. Since the dual of $L^{p'}(\Omega)$ is isomorphic to $L^{p}(\Omega)$, there is some $f_1$ in $L^{p}(\Omega)$, such that $$ F(g) = \int_\Omega f_1g, \quad \forall g\in L^{p'}(\Omega). $$ In particular, for every $\phi$ in $C^\infty_c(\Omega)$ we have that $$ \int_\Omega f\phi = \tilde f(\phi) = F(\phi) = \int_\Omega f_1\phi, $$ so $$ \int_\Omega (f-f_1)\phi = 0, $$ and we conclude that $f=f_1\in L^{p}(\Omega)$.