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

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

• What about Hahn Banach Theorem? Commented Dec 5, 2020 at 17:52
• Continuous linear maps defined on a dense subspace, and taking values in a complete space, can be extended uniquely to a continuous map.
– Ruy
Commented Dec 5, 2020 at 21:30
• 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? Commented Dec 6, 2020 at 0:16

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