# $\int \hat{f} g = 0$ for all test functions $f$ implies $g=0$

Let $$g\in\mathcal{C}^0(\mathbb{R},\mathbb{C})$$ be a bounded continuous function such that $$\int_{\mathbb{R}}\hat{f}g = 0$$ for all test functions $$f\in\mathcal{C}^{\infty}_c(\mathbb{R},\mathbb{C})$$, where $$\hat{f}$$ denotes the Fourier transform of $$f$$.

[The integral $$\int_{\mathbb{R}}\hat{f}g$$ is well-defined for all $$f\in\mathcal{C}^{\infty}_c(\mathbb{R},\mathbb{C})$$ as then $$\hat{f}$$ is Schwartz, and thus rapidly decreasing (and $$g$$ is bounded).]

I want to show that then $$g$$ has to be identically zero. (In particular, for my application, I only need to get $$g(0)=0$$).

Similar questions have already been answered (see e.g. If $f\in L^1_{loc}(\mathbb{R})$ and $\int f\varphi=0$ for all $\varphi$ continuous with compact support, then $f=0$ a.e. or If $\int fg = 0$ for all compactly supported continuous g, then f = 0 a.e.?), but here I deal with the Fourier transform of a test function, so I'm not sure how to tackle this problem.

Any help is greatly appreciated, thanks in advance.

• Every Schwartz function can be written as the Fourier transform of some Schwartz function. May 8, 2023 at 14:42

Let $$\phi$$ be a smooth compactly supported bump function such that $$\phi(x)=1$$ for $$|x|\leq 1$$. For each $$\epsilon>0$$, define $$\phi_{\epsilon}(x)=\phi\left(\epsilon x\right)$$. Then, we have $$\widehat{\phi_\epsilon}(\xi)=\frac{1}{\epsilon}\widehat{\phi}\left(\frac{\xi}{\epsilon}\right)$$. Also, we have $$\int_{\Bbb{R}}\widehat{\phi}(\xi)\,d\xi=\phi(0)=1$$ by Fourier inversion, so we have that $$\{\widehat{\phi_{\epsilon}}\}_{\epsilon>0}$$ is an approximate identity as $$\epsilon\to 0^+$$.

For any Schwartz function $$f$$, let’s define $$f_\epsilon=f\cdot \phi_\epsilon$$. Then, $$f_\epsilon$$ is smooth, compactly supported, and as $$\epsilon\to 0^+$$, we have \begin{align} \widehat{f_{\epsilon}}&=\widehat{f}*\widehat{\phi_{\epsilon}}\to \widehat{f}, \end{align}

where the convergence holds in any $$L^p$$ space with $$p\in [1,\infty)$$. In particular, it holds for $$p=1$$. Therefore, for any $$g\in L^{\infty}$$, we have \begin{align} \left|\int_{\Bbb{R}}\hat{f}g\right|&=\left|\int_{\Bbb{R}}\hat{f}g-\int_{\Bbb{R}}\widehat{f_{\epsilon}}g\right|\leq \|g\|_{L^{\infty}}\|\hat{f}-\widehat{f_{\epsilon}}\|_{L^1}, \end{align} and we just showed the RHS vanishes as $$\epsilon\to 0^+$$, so $$\int_{\Bbb{R}}\hat{f}g=0$$. Since we showed this for all Schwartz functions $$f$$, we deduce that $$\int_{\Bbb{R}}fg=0$$ for all Schwartz functions $$f$$. Thus, by the usual methods, $$g=0$$ a.e. If you now assume $$g$$ is continuous, then of course $$g=0$$ identically.

• Well, this is definitely easier than what I did. Thanks! May 17, 2023 at 6:37

After a lot of thinking, I think I found a way to prove what I wanted (that is, conclude that $$g(0)=0$$), whenever $$g$$ is differentiable with $$g'$$ bounded.

[Notation: I use $$\hat f$$ or $$\mathcal{F}f$$ interchangeably for the Fourier transform of $$f$$, defined here as $$\xi\longmapsto\int_{\mathbb{R}}f(t)e^{-it\xi}\mathrm{d}\nu(t)$$, where $$\nu = \frac{1}{\sqrt{2\pi}}\lambda$$ is a weighted Lebesgue measure.]

For any $$f\in\mathcal{C}^\infty_c(\mathbb{R},\mathbb{C})$$, its Fourier transform $$\hat f$$ is Schwartz, and then so is $$\xi\longmapsto (1+\xi^2)\hat{f}(\xi)$$. By standard properties of the Fourier transform, notice that the latter can be written as $$\mathcal{F}(f-f'')$$. Therefore we have

$$0 = \int_{\mathbb{R}}\hat fg\ \mathrm{d}\lambda=\int_{\mathbb{R}}\mathcal{F}(f-f'')(\xi)\frac{g(\xi)}{1+\xi^2}\mathrm{d}\lambda(\xi).$$

Since $$g$$ is continuous and bounded, the Fourier transform of $$h:\xi\longmapsto\frac{g(\xi)}{1+\xi^2}$$ is well-defined as $$h\in L^1(\mathbb{R},\mathbb{C})$$. Notice that in addition, $$h\in L^2(\mathbb{R},\mathbb{C})$$, so $$\hat{h}\in L^2(\mathbb{R},\mathbb{C})$$. Using the product property of $$\mathcal F$$, $$0 = \int_{\mathbb{R}} \mathcal{F}(f-f'')h\ \mathrm{d}\lambda=\int_{\mathbb{R}} (f-f'')\mathcal{F}h\ \mathrm{d}\lambda.\quad(1)$$

Since $$g$$ is differentiable with bounded derivative (by assumption), we have that $$h$$ is differentiable as well, with $$h'\in L^1(\mathbb{R},\mathbb{C})$$. As $$|h(\xi)|\xrightarrow{|\xi|\to\infty}0$$, by integration by parts, $$\widehat{h'}=[\xi\mapsto i\xi\hat{h}(\xi)]$$. It is easily checked that $$h'\in L^2(\mathbb{R},\mathbb{C})$$ as well, therefore $$[\xi\mapsto (1+|\xi|)|\hat{h}(\xi)|]\in L^2$$. This implies that $$\hat{h}\in L^1(\mathbb{R},\mathbb{C})$$ since by Cauchy-Schwartz, $$\int_{\mathbb{R}}|\hat{h}|\mathrm{d}\lambda = \int_{\mathbb{R}}(1+|\xi|)|\hat{h}(\xi)|\frac{1}{1+|\xi|}\mathrm{d}\lambda(\xi) \leq \left\lVert(1+|\cdot|)|\hat{h}|\right\rVert_{L^2}\left\lVert\tfrac{1}{1+|\cdot|}\right\rVert_{L^2} < \infty.$$ By the Fourier inversion formula for $$L^2$$ (see e.g. Rudin's RCA, Theorem 9.13), denoting $$h_n: t\longmapsto\int_{-n}^{n}\hat{h}(\xi)e^{it\xi}\mathrm{d}\nu(\xi)$$, we have that $$h_n\longrightarrow h$$ in the $$L^2$$ sense. In particular, it means that there is a subsequence $$(h_{n_k})_k$$ such that $$h_{n_k}\longrightarrow h$$ almost everywhere. Since by assumbtion, $$h$$ is continuous, we have that $$h_n(0)\longrightarrow h(0) = g(0)$$. Since $$\hat{h}$$ is $$L^1$$, by the DCT, we have in particular that $$\int_{\mathbb{R}}\hat{h}\ \mathrm{d}\lambda=\lim_{n\to\infty}h_n(0)=g(0).$$

Now, let $$(\phi_n)_n\subset\mathcal{C}^\infty_c(\mathbb{R},\mathbb{C})$$ a sequence of smooth functions such that $$\operatorname{supp}(\phi_n)=[-n-1{,}n+1]$$ and $$\phi_n\equiv 1$$ on $$[-n,n]$$. Since $$\phi_n\longrightarrow 1$$ pointwise and $$\hat{h}$$ is $$L^1$$, by the DCT, we have $$\lim_{n\to\infty}\int_{\mathbb{R}}\phi_n\hat{h}\ \mathrm{d}\lambda=\int_{\mathbb{R}}\hat{h}\ \mathrm{d}\lambda = g(0).$$ On the other hand, for any $$n$$, we have that $$\phi_n''$$ has support $$[-n-1{,}-n]\cup[n{,}n+1]$$, and the sequence $$(\phi_n)_n$$ can be chosen such that there exists some $$C>0$$ with $$|\phi_n''| for all $$n$$. Thus, since $$\hat{h}$$ is $$L^1$$, $$\lim_{n\to\infty}\left\lvert\int_{\mathbb{R}}\phi_n''\hat{h}\ \mathrm{d}\lambda\right\rvert \leq C\lim_{n\to\infty}\int_{[-n-1{,}-n]\cup[n{,}n+1]}\lvert\hat{h}\rvert\ \mathrm{d}\lambda = 0.$$

By (1), this allows us to conclude that $$0 = \lim_{n\to\infty}\int_{\mathbb{R}} (\phi_n-\phi_n'')\mathcal{F}h\ \mathrm{d}\lambda = \lim_{n\to\infty}\int_{\mathbb{R}}\phi_n\hat{h}\ \mathrm{d}\lambda - \lim_{n\to\infty}\int_{\mathbb{R}}\phi_n''\hat{h}\ \mathrm{d}\lambda = g(0),$$ which is what we wanted.

Feel free to comment if you find anything incorrect.