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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. $\sup\limits_{g\in L^{p'}(\Omega)}\frac{\lvert\tilde{f}(g)\rvert}{\lVert g \rVert}\le 1$.

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  • $\begingroup$ What about Hahn Banach Theorem? $\endgroup$
    – ProfOak
    Commented Dec 5, 2020 at 17:52
  • $\begingroup$ Continuous linear maps defined on a dense subspace, and taking values in a complete space, can be extended uniquely to a continuous map. $\endgroup$
    – Ruy
    Commented Dec 5, 2020 at 21:30
  • $\begingroup$ But when you use Hahn Banach, you consider the restriction $\tilde{f}:C_0^{\infty}(\Omega)\to \mathbb{R}$ (and in particular for this restriction we have $\lVert \tilde{f}\rVert$ so there is an extensison $h\in(L^{p'})^*$ with norm $1$ as well but how do you know that $h$ has the same expression than $\tilde{g}$ on the whole space? $\endgroup$
    – roi_saumon
    Commented Dec 6, 2020 at 0:16

1 Answer 1

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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 $f=f_1\in L^{p}(\Omega)$.

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