$f \in L^1(R)$, Fourier transform $\hat{f} \in L^2(R)$ implies $f \in L^2(R)$

Question

If I have $f \in L^1(\mathbb{R})$ and that the Fourier transform $\hat{f} \in L^2(\mathbb{R})$, how can I show that $f \in L^2(R)$? I was thinking about ways to apply Fourier inversion, but could not think of anything. What is a good approach to take here?

Answer 1

This solution is going to seem a bit abstruse, but there aren't a lot of elementary ways to get at this. Using Fourier's heat solution is one of those ways.

If $f,g\in L^1$, then $$ \int_{\mathbb{R}}\hat{f}(u)g(u)du = \int_{\mathbb{R}}f(u)\hat{g}(u)du. $$ That's an elementary fact which is often overlooked, and one of the few Fourier transform theorems that requires almost no conditions. For example, \begin{align} \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat{f}(u)e^{iux}e^{-tu^2}du & = \int_{\mathbb{R}}f(u)\frac{e^{-(u-x)^2/4t}}{\sqrt{4\pi t}}du \end{align} (This is Fourier's solution to the heat equation $u_t = u_{xx}$ with $u(0,x)=f(x)$.) The left side converges to $(\hat{f})^{\vee}$ in $L^2$ as $t\downarrow 0$ because $\hat{f}\in L^2$ is assumed (which forces $\hat{f}(u)e^{-tu^2}\rightarrow \hat{f}$ in $L^2$ and $t\downarrow 0$,) and because the $L^2$ Fourier transform is unitary by the Plancherel Theorem. The right side converges to $f$ in $L^1$ as $t\downarrow 0$ because $f\in L^1$ and because of properties of the Heat Kernel. Hence, there is a subsequence $\{ t_n \}$ tending downward to $0$ along which the left side converges pointwise a.e. to $(\hat{f})^{\vee}$, and along which the right side converges pointwise a.e. to $f\in L^1$. Therefore $(\hat{f})^{\vee}=f$ a.e., which proves that $f\in L^2$ because $\hat{f}\in L^2$ forces $(\hat{f})^{\vee} \in L^2$ by the Plancherel Theorem.

Answer 2

I think the answer should be that if $f,g \in L^1$ $$\langle \hat{f},g \rangle = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x) e^{-2i \pi y x}dx \overline{g(y)} dy =\int_{-\infty}^\infty f(x)\int_{-\infty}^\infty \overline{e^{2i \pi y x} g(y)} dy dx = \langle f,\overset{\vee}{g} \rangle$$ (ie. the adjoint of the Fourier transform is its complex conjugate)

and $\langle \hat{f},g \rangle = \langle f,\overset{\vee}{g} \rangle$ stays true for $\hat{f},g\in L^2$ because $L^1 \cap L^2$ is dense in $L^2$.

Thus $$\|\hat{f}\|_{L^2}^2 =\langle \hat{f}, \hat{f}\rangle= \langle f, \overset{\vee}{\hat{f}}\rangle= \langle f, f\rangle= \|f\|_{L^2}^2$$ Where we used the Fourier inversion theorem $f = \overset{\vee}{\hat{f}}$ (the equality is in $L^2$).

Qed. $\hat{f} \in L^2 \implies f \in L^2$.

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The only alternative I see (avoiding the Fourier inversion theorem) is proving that $f \not \in L^2 \implies \hat{f} \not \in L^2$ but I'm affraid we'll need to say the (linear combination of) Gaussians (or something equivalent with a known Fourier and inverse Fourier transform) are dense in $L^2$, which is similar to the proof of the Fourier inversion theorem.

Answer 3

Let $\mathcal{F}^{-1}(\hat{f})=g\in L^2(\mathbb{R})$. We will show $f=g$ a.e. Now we will show that $$ \int u\hat{v}=\int\hat{u}v$$ for all $u\in L^1(\mathbb{R})$ and $v\in\mathcal{S}(\mathbb{R})$. This is clearly true for both $u,v\in\mathcal{S}(\mathbb{R})$ so since $\mathcal{S}(\mathbb{R})$ is dense in $L^1(\mathbb{R})$, we can take a sequence $u_n\in\mathcal{S}(\mathbb{R})$ such that $u_n\to u$ and $u_n(x)\le u(x)$ a.e. We know that $$ \int u_n\hat{v}=\int\hat{u_n}v.$$ Since $v$ is Schwartz, it is continuous and bounded so $u\hat{v}\in L^1(\mathbb{R})$ since $\hat{v}$ is Schwartz and $u$ is $L^1(\mathbb{R})$. We also have $\hat{u}v\in L^1(\mathbb{R})$ since $\hat{u}$ is bounded and continuous while $v$ is in $L^1(\mathbb{R})$. Therefore, we can use Dominated Convergence to see that $$ \int u_n\hat{v}\to\int u\hat{v}$$ and $$ \int\hat{u_n}v\to\int\hat{u}v$$ so we must have $$ \int u\hat{v}=\int\hat{u}v.$$ Next, we will show that $$ \int f\hat{\varphi}=\int g\hat{\varphi}$$ for all $\varphi\in\mathcal{S}(\mathbb{R})$. This is because $$ \int f\hat{\varphi}=\int\hat{f}\varphi=\int\hat{g}\varphi=\int g\hat{\varphi}.$$ Define $L^1_{loc}(\mathbb{R})$ to be the space of locally integrable functions. We will show that if $f\in L^1_{loc}(\mathbb{R})$ and $$ \int f\varphi=0$$ for all $\varphi\in\mathcal{S}(\mathbb{R})$ then $f=0$ a.e. First, we will show this for $f\in L^1(\mathbb{R})$. Since $\mathcal{S}(\mathbb{R})$ is dense in $L^1(\mathbb{R})$, we can take a sequence $\psi_n\in\mathcal{S}(\mathbb{R})$ such that $\psi_n\to \frac{f}{|f|+1}$ and so that $|f\psi_n|\le|f|$. We also have $f\psi_n\to\frac{f^2}{|f|+1}$ so by Dominated Convergence we have $$ \int \frac{f^2}{|f|+1}=0$$ since $$ \int f\psi_n=0$$ for all $n$. However, this implies that we have $f=0$ a.e. To extend this to all $f\in L^1_{loc}(\mathbb{R})$, consider $f'=fg$ where $g=1$ on $[-n,n]$ and is continuous with compact support. Note that $f'$ is in $L^1(\mathbb{R})$ and satisfies the hypothesis on $[-n,n]$ so we must have $f=0$ on $[-n,n]$. Since this holds for any $n$, we can see that $f=0$ a.e. Finally, note that since $f\in L^1(\mathbb{R})\subset L^1_{loc}(\mathbb{R})$ and $g\in L^2(\mathbb{R})\subset L^1_{loc}(\mathbb{R})$, we have $f-g\in L^1_{loc}(\mathbb{R})$. Also, $$ \int f\hat{\varphi}=\int g\hat{\varphi}\Rightarrow \int (f-g)\hat{\varphi}=0$$ for all $\varphi\in\mathcal{S}(\mathbb{R})$. Therefore, by our previous result, we must have $f-g=0$ a.e. Therefore, we must have $f\in L^2(\mathbb{R})$.