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For a function $f\in L^1(\mathbb{R})$, its Fourier transform is defined as $$\hat{f}(y)=\int_{-\infty}^\infty f(x)e^{-ixy}dx$$

For a function $f\in L^2(\mathbb{R})$, its Fourier transform is defined as the unique continuous mapping $g:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ that extends the mapping $h:S\rightarrow L^2(\mathbb{R})$, where $S$ is the Schwartz class, and the Fourier transform of a function in the Schwartz class is defined as in the first paragraph. (We may assume that this continuous mapping $g$ exists and is unique.)

Let $f$ be in both $L^1(\mathbb{R})$ and $L^2(\mathbb{R})$. Show that the two definitions coincide.

By the second definition, we must find functions $g_1,g_2,\ldots\in S$ such that $\|g_n-f\|_2\rightarrow 0$ as $n\rightarrow\infty$. Then $\hat{f}$ is the unique function such that $\|\hat{g_n}-\hat{f}\|_2\rightarrow 0$ as $n\rightarrow\infty$. So we must prove that

$$\int_{-\infty}^\infty g_n(x)e^{-ixy}dx-\int_{-\infty}^\infty f(x)e^{-ixy}dx\rightarrow 0\text{ as } n\rightarrow\infty$$ That is, $$\int_{-\infty}^\infty (g_n(x)-f(x))e^{-ixy}dx\rightarrow 0\text{ as } n\rightarrow\infty$$

But why is this true? We only know that $$\int_{-\infty}^\infty (g_n(x)-f(x))^2dx\rightarrow 0\text{ as } n\rightarrow\infty$$

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Let us, to distinguish between the two Fourier transforms, denote the one by $\mathscr{F}_1$,

$$\mathscr{F}_1(f)(y) = \int_\mathbb{R} f(x)e^{-ixy}\,dx\tag{1}$$

for $f\in L^1(\mathbb{R})$, and the other by $\mathscr{F}_2$,

$$\mathscr{F}_2(g) = \lim_{n\to\infty} \mathscr{F}_1(g_n)\tag{2}$$

for $g\in L^2(\mathbb{R})$ where $g_n$ is a sequence of functions in $S$ that converges to $g$ in $L^2$, and the limit in $(2)$ is the $L^2$-limit.

For $f \in L^1 \cap L^2$, find a sequence $(g_n)$ in $S$ such that not only $\lVert g_n - f\rVert_2 \to 0$ but also $\lVert g_n - f\rVert_1 \to 0$.(1)

Then you can conclude

$$\left\lVert \mathscr{F}_1(g_n) - \mathscr{F}_1(f)\right\rVert_\infty \leqslant \lVert g_n - f\rVert_1 \to 0,$$

i.e. the Fourier transforms of the $g_n$ converge uniformly to the $L^1$ Fourier transform of $f$, and by definition you have

$$\mathscr{F}_1(g_n) \xrightarrow{L^2} \mathscr{F}_2(f),$$

and there is a subsequence of the $\mathscr{F}_1(g_n)$ that converges pointwise almost everywhere to $\mathscr{F}_2(f)$ (well, we know it converges uniformly, so the entire sequence converges pointwise), so you have $\mathscr{F}_1(f) = \mathscr{F}_2(f)$ almost everywhere (and you can choose $\mathscr{F}_1(f)$ as a representative, so then you have equality everywhere). In that sense, the two definitions coincide, for $f \in L^1\cap L^2$, the $L^1$-Fourier transform of $f$ is a representative of the $L^2$-Fourier transform of $f$.

If you choose a sequence $(g_n)$ in $S$ such that $\lVert g_n - f\rVert_2 \to 0$, but not $\lVert g_n -f \rVert_1 \to 0$, then by the above you have convergence $\hat{g}_n \xrightarrow{L^2} \hat{f}$, and that implies that for a subsequence $g_{n_k}$, you have pointwise convergence $\hat{g}_{n_k}(y) \to \hat{f}(y)$ almost everywhere, but there is no guarantee (known to me) that you have pointwise a.e. convergence for the full sequence, nor that you have pointwise convergence everywhere for any subsequence. But it would be difficult at least, I think, to come up with a concrete example of a function $f\in L^1\cap L^2$ and a sequence $g_n$ of Schwartz functions converging to $f$ in $L^2$ such that you don't have pointwise convergence almost everywhere or even everywhere for the full sequence.


(1) That is always possible, let $f_m(x) = f(x)\cdot \chi_{[-m,m]}(x)\cdot \chi_{\{ \lvert f(y)\rvert \leqslant m\}}(x)$. Then $f_m \to f$ both in $L^1$ and $L^2$. $f_m$ is a bounded function with compact support, so convolving it with a compactly supported approximation of the identity produces compactly supported smooth functions converging to $f_m$ in both $L^1$ and $L^2$.

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  • $\begingroup$ Thanks Daniel. Your approach looks interesting, but meanwhile I also wonder why the last convergence in my original post holds. Any ideas on that? $\endgroup$ – PJ Miller Nov 9 '13 at 22:07
  • $\begingroup$ If you choose a sequence $g_n$ that converges to $f$ in both $L^1$ and $L^2$, the $L^1$ convergence implies that the Fourier transforms of the $g_n$ converge uniformly to that of $f$. If the sequence $g_n$ converges to $f$ only in $L^2$, you don't necessarily have pointwise convergence $\int (g_n(x)-f(x))e^{-ixy}\,dx \to 0$. But since you have $L^2$ convergence, for a subsequence $g_{n_k}$, you have pointwise convergence almost everywhere. $\endgroup$ – Daniel Fischer Nov 9 '13 at 22:32
  • $\begingroup$ I've been thinking about why $\left\lVert \mathscr{F}_1(g_n) - \mathscr{F}_1(f)\right\rVert_\infty \leqslant \lVert g_n - f\rVert_1$, but can't see why that's true. $\endgroup$ – PJ Miller Nov 9 '13 at 22:43
  • $\begingroup$ You have $$\lvert \hat{f}(y)\rvert = \left\lvert \int f(x)e^{-ixy}\,dx \right\rvert \leqslant \int \lvert f(x)\rvert\,dx = \lVert f\rVert_1,$$ the absolute value of an $L^1$ Fourier transform is bounded by the $L^1$ norm (possibly with a normalisation factor, depends on where you put the $2\pi$ in the FT). So since $\mathscr{F}_1(g_n) - \mathscr{F}_1(f) = \mathscr{F}_1(g_n-f)$, the estimate follows. $\endgroup$ – Daniel Fischer Nov 9 '13 at 22:47

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