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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 ?

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    $\begingroup$ You forgot to specify who $u$ is. $\endgroup$ – Giuseppe Negro Jan 8 '16 at 10:31
  • $\begingroup$ "Show that $f $ is bounded". This is part of your assumption. Do you mean "show that $u $ us bounded"? $\endgroup$ – PhoemueX Jan 8 '16 at 17:31
  • $\begingroup$ Yes we have to show that $u$ is bounded $\endgroup$ – M.LTA Jan 8 '16 at 17:33
  • $\begingroup$ @Giuseppe Negro $ u $ is a primitive of $ f it is said. $\endgroup$ – M.LTA Jan 10 '16 at 20:04
  • $\begingroup$ Try expressing $u$ as convolution between $f$ and a Heaviside step function: $$u(x)=\int_{-\infty}^x f(s)\, ds=f\ast H, $$ where $H(t)=\mathbb{1}_{t\ge 0}$. $\endgroup$ – Giuseppe Negro Jan 10 '16 at 20:09
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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 ?

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