I' asked to show the following statement: Let $f \in L^2(\mathbb{R}^n)$ and $T \in S^*(\mathbb{R}^n)$. Then there exists $F \in L^2(\mathbb{R}^n)$ such that $\hat{T}_f=T_F$ (where the hat denotes the Fourier transform). My first problem: I suppose that by $T_f$ we mean the tempered distribution induced by $f$. However, I couldn't find any definition in my notes. Could something else be meant, for example a more general tempered distribution?
Assuming $T_f$ denotes the tempered distribution induced by $f$, I approached the problem as follows:
Let $f \in L^2(\mathbb{R}^n)$. Since $C_0^\infty$ is dense in $L^2$, we can approximate the function $f \in L^2(\mathbb{R}^n)$ by $f_j \in C_0^\infty$ . Then, for all $u \in S(\mathbb{R}^n)$ we have that \begin{equation} (\hat{T}_{f_j})(u) = \int_{\mathbb{R}^n} f_j \hat{u} = \int_{\mathbb{R}^n} \hat{f_j} {u} = T_{\hat{f}_j}(u). \end{equation} wher the first equality follows from the definition of $\hat{T}_{f_j}$ and the second equality holds since $\int \hat{g} f = \int g \hat{f} $ for all $f,g \in L^1(\mathbb{R}^n)$, so in particular for $f_j \in C_0^\infty(\mathbb{R}^n)$ and $u \in S(\mathbb{R}^n)$.
By Hölder, we can now see that \begin{equation} |\hat{T}_{f_j}(u)-\hat{T}_f(u)| \leq |\int_{\mathbb{R}^n} (f_j-f)\hat{u}| \leq ||f_j-f||_{L^2(\mathbb{R}^n} ||\hat{u}||_{L^2(\mathbb{R}^n)} = ||f_j-f||_{L^2(\mathbb{R}^n)} ||{u}||_{L^2(\mathbb{R}^n} \end{equation} where we used Plancherel in the last step. Now, since $f_j\rightarrow f$ in $L^2(\mathbb{R}^n)$, we can see that the expression above tend to $0$ as $j \rightarrow \infty$.
On the other hand, since $f_j \rightarrow f$ in $L^2(\mathbb{R}^n)$, we also have that $\hat{f_j} \rightarrow \hat{f}$ in $L^2(\mathbb{R}^n)$ by the continuity of the Fourier transform. Thus, \begin{equation} |T_{\hat{f}_j}(u)-T_\hat{f}(u)| \leq |\int \hat{f_j} u - \int \hat{f}u| \leq ||\hat{f}_j-\hat{f}||_{L^2(\mathbb{R}^n)} ||u||_{L^2(\mathbb{R}^n)} \end{equation} which goes to $0$ as $j$ goes to $\infty$.
Hence, $\hat{T}_f=T_{\hat{f}}$ where $\hat{f} \in L^2(\mathbb{R}^n)$. Since the Fourier transform in $L^2(\mathbb{R}^n)$ is bijective, we know that $\hat{f} \in L^2(\mathbb{R}^n$. Hence, the required $F$ is given by $\hat{f}$.
Is this proof correct? I'm just a bit unsure since a hint was given that asked to use the Riesz Representation Theorem. I don't know where I would be supposed to use this theorem, is there some mistake in my proof?