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Let $\varphi \in \mathcal{D}(\mathbb{R})$ and let $\theta \in \mathcal{D}(\mathbb{R})$ such $\theta=1$ au voisinage de 0. My question is: please, how we prouve that there exist $\psi \in \mathcal{D}(\mathbb{R})$ such as $$ \varphi(x)= \varphi(0) \theta(x)+ x \psi(x) $$ with the indication to considerate $\psi(x)= \displaystyle\int_0^1 \varphi'(tx) dt$?

Thank you for the help, because i have no idea to the proof.

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  • $\begingroup$ Do the integral (a simple variable change), substitute and check the equality. $\endgroup$ – Rafa Budría Feb 20 '17 at 18:13
  • $\begingroup$ sorry, i don't understand. Can you more explain me please $\endgroup$ – user415040 Feb 20 '17 at 18:24
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Let's clean a little:

$$\psi(x)= \displaystyle\int_0^1 \varphi'(tx)\mathbb dt$$

Now, let $z=tx$, then $\mathbb dz=x\mathbb dt$, $z=0$ for $t=0$ and $z=x$ for $t=1$

$$\psi(x)= \frac{1}{x}\int_0^x \varphi'(z)\mathbb dz$$

$$\psi(x)=\frac{1}{x}[\varphi(z)]_0^x=\frac{1}{x}(\varphi(x)-\varphi(0))$$

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