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 $2=2'$.