# Fourier transform of a distribution null on [-1,1]

Here is an interesting problem :

Let $f \in \mathcal{C}^0 ( \mathbb{R})$ bounded, and $T_f \in \mathcal{S}'(\mathbb{R})$ defined by $\displaystyle \langle T_f, \phi \rangle = \int_{\mathbb{R}} f(x)\phi (x) \mathrm{d}x$. Suppose that $\langle \hat{T_f}, \phi \rangle =0$ for every test function $\phi$ compactly supported in $]-1,1[$. Let $u$ be a primitive of $f$ (an antiderivative) Show that $u$ is bounded and compute the support of $\hat{u}$ (as a distribution)

What I've done so far. First of all, the fact that $f$ is bounded ensure that we really have $T_f \in \mathcal{S}'(\mathbb{R})$. Then I set : $$\langle u , \phi \rangle = \int_{\mathbb{R}} \left( \int_{0}^x f(t) \mathrm{d}t \right) \phi (x)\mathrm{d}x$$ Thus I take $\phi$ such that $\phi (x)=1$ for $|x|<1/n$ and $\phi (x)=0$ elsewhere. Then, using $\langle \hat{T_f}, \phi \rangle =0$ I computed : $$\langle \hat{T_f}, \phi \rangle = \langle T_f, \hat{\phi}\rangle = \int_{\mathbb{R}} f(x)\int_{\mathbb{R}} \phi(t)e^{-ixt} \mathrm{d}t \mathrm{d}x = 2 \int_{\mathbb{R}} f(x)\frac{\sin \left( \frac{x}{n} \right)}{x}\mathrm{d}x$$ So that : $$\int_{- \infty}^{+ \infty} f(x) \frac{\sin \left( \frac{x}{n}\right)}{x} \mathrm{d}x =0$$ for every $n \geqslant 0$

Still it doesn't prove much. So how can I continue and conclude ? Am I on the right direction ?

• You forgot to specify who $u$ is. – Giuseppe Negro Jan 8 '16 at 10:31
• "Show that $f$ is bounded". This is part of your assumption. Do you mean "show that $u$ us bounded"? – PhoemueX Jan 8 '16 at 17:31
• Yes we have to show that $u$ is bounded – M.LTA Jan 8 '16 at 17:33
• @Giuseppe Negro $u$ is a primitive of $f it is said. – M.LTA Jan 10 '16 at 20:04 • Try expressing$u$as convolution between$f$and a Heaviside step function: $$u(x)=\int_{-\infty}^x f(s)\, ds=f\ast H,$$ where$H(t)=\mathbb{1}_{t\ge 0}$. – Giuseppe Negro Jan 10 '16 at 20:09 ## 1 Answer The answer above is not mine : it answers the problem precisely with not any shadow point. Nevertheless, I found this proof not natural, and by reading it you will probably understand what I mean. First,$f$is seen as a tempered distribution, as$\hat f$For the proof, we set$\chi$a test function such that$\operatorname{supp} (\chi) \subset (-1,1)$and$\chi (x)=1$for$x \in \left( -\dfrac{1}{2}, \dfrac{1}{2} \right)$. Note that$\hat f \chi =0$. Then remark that$h: \xi \mapsto \frac{(1-\chi (\xi))}{i \xi}$with$h(0)=0$is$\mathcal{C}^{\infty}$. Then, we set :$$u(x)=\mathcal{F} ^{-1} (hf)$$ Then : $$\hat{u'}=i \xi \hat u = (1- \chi) \hat f = \hat f$$ so that$u$is an antiderivative of$f$. This also shows that$\operatorname{supp} \hat u = \operatorname{supp} \hat f$If$u(x) \geqslant A$using that$u$is$\|f\|_{L^{\infty}}$- Lipschitz, we get : $$f(y) \geqslant A - \|f\|_{L^{\infty}}$$ for every$y \in[x-1,x+1]$. Thus : $$\int_{\mathbb{R}} u(y) \chi (x-y)\mathrm{d}y \geqslant (A - \|f\|_{L^{\infty}}) \int_{\mathbb{R}} \chi (y) \mathrm{d}y$$ All we have to prove is that : $$g : \lambda \longmapsto \int_{\mathbb{R}} u(y) \chi (y-\lambda) \mathrm{d}y$$ is bounded. Write : $$g(\lambda) = \langle u, \chi ( \cdot - \lambda )\rangle = \frac{1}{2 \pi} \langle u, \widehat{\hat {\chi}} ( \cdot - \lambda)\rangle = \frac{1}{2 \pi} \langle u, \widehat{e^{-i \lambda \xi}\hat {\chi}} ( \xi)\rangle = \frac{1}{2 \pi} \langle \hat{u}, e^{-i \lambda \xi}\hat {\chi} ( \xi)\rangle$$ We obtain : $$g(\lambda) = \frac{1}{2 \pi} \left \langle f, \mathcal{F} \left( e^{-i \lambda \xi} \frac{\hat{\chi} (\xi) (1- \chi (\xi))}{i \xi}\right)\right \rangle$$ Since the function$\psi : \xi \mapsto \frac{\hat{\chi} (\xi) (1- \chi (\xi))}{i \xi}$belongs to the Schwartz space and that : $$\| \mathcal{F}(e^{i \lambda \xi} \psi)\|_{L^1}=\|\hat \psi\|_{L^1}$$ we can write : $$|g(\lambda)| \leqslant \|f\|_{L^{\infty}} \|\hat \psi\|_{L^1}$$ So that$g$is bounded and so$u\$ is.

Something more natural to submit ? Or maybe motivations of the proof ?