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Consider a distribution $T \in D'(\mathbb{R})$ such as (E) : $T' + gT = 0$ with $g \in D(\mathbb{R})$. Could you prove that $T$ is a strong solution of (E) ? I know that we must use the fondamental theorem of calculus with distribution but i don't know how to conclude...

Thanks for answers

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Do you know that distributions which have a derivative equal to $0$ on $\mathbb R$ are constant? –  Davide Giraudo Mar 10 '12 at 18:26
    
Yes i know, but i don't see how to use That... –  Adrien Boulanger Mar 10 '12 at 18:38
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Take $G$ a primitive of $g$, and multiply by $e^G$, and show that $e^GT$ has a derivative which is $0$. –  Davide Giraudo Mar 10 '12 at 18:40
    
Ok i'm stupid like the real case... :-) thanks –  Adrien Boulanger Mar 10 '12 at 18:42

1 Answer 1

up vote 1 down vote accepted

Let $G$ a primitive of $g$, which exist since $g$ is continuous. Let $\varphi$ a test-function. We have, if $T$ is a solution of $T'+gT=0$ that \begin{align*} \langle (e^GT)',\varphi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)} &=-\langle e^GT,\varphi'\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)} \\ &=-\langle T,e^G\varphi'\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}\\ &=-\langle T,(e^G\varphi)'-G'e^G\varphi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}\\ &=-\langle T',e^G\varphi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}+ \langle T,ge^G \varphi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}\\ &=\langle -gT,e^G\varphi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}+ \langle T,ge^G \varphi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}=0. \end{align*} Now we use the fact that if a distribution on $\mathbb R$ has a derivative equal to $0$, it can be represented by a constant function (see here for example). So $e^GT$ can be represented by the constant $C$ and so $T$ is also a strong solution of (E).

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