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
Best Answer
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