Question on distributions theory

i have the following question: let $\Omega$ an open in $\mathbb{R}^n$, and let $T \in \mathcal{D}'(\Omega)$.

We suppose that there exists an positive constant $C$ such that $$\forall \varphi \in \mathcal{D}(\Omega): |\langle T,\varphi\rangle| \leq C \sqrt{ \displaystyle\int_{\Omega} |\varphi(x)|^2 dx }$$

The question is to prouve that there exists $f \in L^2(\Omega)$ such that $$\forall \varphi \in \mathcal{D}(\Omega): \langle T,\varphi\rangle = \displaystyle\int_{\Omega} f(x) \varphi(x) dx.$$

My problem is that i have no idea for an probable solution. Can you give me somes indocations please.

There are two things going on, which is a bit confusing in the case $p=2$ because they look like the same thing. The first thing is that $T$ can be extended from a continuous linear functional on $D(\Omega)$ to a continuous linear functional on $L^2(\Omega)$. To define it, you just define $T(f)=\lim_{n \to \infty} T(f_n)$ where $f_n$ are in $D$ and converge in $L^2$ to $f$. You need the given bound to show that this does not depend on the choice of $f_n$.
The second thing is that continuous linear functionals on $L^2$ are given by integration against a function in $L^2$. This is the content of the Riesz representation theorem in the case of $L^2$. If instead $T$ were given to act continuously on $L^3$, then you would find that $T$ is represented by integration against a function in $L^{3'}$, where $\frac{1}{3}+\frac{1}{3'}=1$ so that $3'=\frac{3}{2}$. $L^2$ is special in a number of ways because $2=2'$.
• Thank you so much for the answer. Please i have two questions: 1. How we use the inequality to prouve that $T(f)=\lim_n T(f_n)$ don't depend to the choice for the sequence $(f_n)$? 2. Please, how we can generalize the question to $L^p$ with $p \neq 2$? – user415040 Mar 19 '17 at 10:16
• @jiji Suppose you have $f_n \to f,g_n \to f$ in $L^2$, then use the bound to check that $T(f_n - g_n) \to 0$. The generalization to $L^p$ is that if it bounded on $L^p$, i.e. $|T(f)| \leq C \| f \|_{L^p}$, then it is represented by integration against a function on $L^{p'}$. – Ian Mar 19 '17 at 15:53
$C_c^\infty(\Omega)$ is dense in $L^2(\Omega)$, so for every $g\in L^2(\Omega)$ you can find a sequence $\{\varphi_n\}$ converging to $g$ in $L^2(\Omega)$. From your inequality you get that $|\langle T,\varphi_n-\varphi_m\rangle|\le C\Vert \varphi_n-\varphi_m\Vert_{L^2}\to 0$ as $n,m\to\infty$. Hence $\{\langle T,\varphi_n\rangle\}$ is a Cauchy sequence and so it has a limit that you call $\langle T,g\rangle$. So you have extended $T$ to $L^2(\Omega)$. You can check that $T$ is linear and continuous and so by Riesz representation theorem you can write it as $\langle T,g\rangle=\int_\Omega fg\,dx$ for some $f\in L^2(\Omega)$.