In my adv. Analysis class we have seen that the Fourier Transform $\mathcal{F}$ is an isometry from $\mathcal{S}(\mathbb{R}^d)$ into itself with respect to the norm $L^2(\mathbb{R}^d)$, where $\mathcal{S}(\mathbb{R}^d)$ is the Schwartz space.
Then we've used the density of $\mathcal{S}(\mathbb{R}^d)$ in $L^2(\mathbb{R}^d)$ (and the fact that $L^2$ is a Banach space) to extend (uniquely) the Fourier Transform to $L^2(\mathbb{R}^d)$. Now we have that $$\mathcal{S}(\mathbb{R}^d) \subset L^1(\mathbb{R}^d) \cap L^2(\mathbb{R}^d) \subset L^2(\mathbb{R}^d)$$ Therefore $L^1(\mathbb{R}^d) \cap L^2(\mathbb{R}^d)$ is dense in $L^2(\mathbb{R}^d)$, and we can use this to evaluate the transform of an $L^2(\mathbb{R}^d)$ function as the ($L^2$) limit of a sequence of transformed $L^1(\mathbb{R}^d) \cap L^2(\mathbb{R}^d)$ functions, that is: $$f \in L^2(\mathbb{R}^d), \; \{f_n\} \subset L^1(\mathbb{R}^d) \cap L^2(\mathbb{R}^d) \text{ s.t. } f_n \xrightarrow{L^2}f \Longrightarrow \hat{f_n} \xrightarrow{L^2} \hat{f}$$
This is useful since we can evaluate the Transform of an $L^1$ function as a Lebesgue integral, and it is much easier to find a convergent sequence in $L^1 \cap L^2$ (to our $L^2$ function) then in $\mathcal{S}$.
Now, this makes sense to me, but when I look for the details, something doesn't add up. In particular:
- who guarantees us that the extendend functional $\mathcal{F}$ over $L^2$ from ${\mathcal{S}}$ will be the same on $L^1 \cap L^2$ as the original Fourier Transform?
This is crucial, since the whole point of this construction is to be able to approximate an $L^2$ transform by well-defined Lebesgue integrals.
We've done this before introducing temperated distributions. So, moreover, how can the introduction of temperated distribution shed light upon this reasoning? Thanks
You want to prove that $F(f)=\hat{f}$ on $L^1 \cap L^2$
Let $f \in L^1 \cap L^2$
Note that exist Schwartz functions $g_n$ such that $||g_n-f||_1 \to 0$ and $||g_n-f||_2 \to 0$
From this we have that $$\sup_{x \in \Bbb{R}^d}|\hat{g_n}(\xi)-\hat{f}(\xi)| \leq ||g_n-f||_1 \to 0$$
Thus $\hat{g_n}(\xi) \to \hat{f}(\xi),\forall \xi$
Also from the definition of fourier transform in $L^2$ and from Plancherel and the fact that $\hat{g_n}$ are Schwartz functions we have that $$||\hat{g_n}-F(f)||_2 \to 0$$
So exists a subsequence $\hat{g_{n_k}} \to F(f)$ a.e
Thus $F(f)=\hat{f}$ a.e