how we can prouve that if $T \in \mathcal{D}'(\mathbb{R})$ and $xT=0$ then $Supp T=\{0\}$? Thank you for help.
Supp of distribution- $\delta$
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functional-analysis
distribution-theory
2 Answers
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Hint: if the support of $\phi\in\mathcal{D}(\Bbb R)$ is included in $\Bbb R\setminus\{0\}$ then $$ \phi= x\,\frac{\phi}{x}, \quad\frac{\phi}{x}\in\mathcal{D}(\Bbb R). $$ Now $$ \langle T,\phi\rangle=\langle T,x\,\frac{\phi}{x}\rangle=\langle x\,T,\frac{\phi}{x}\rangle=0, $$ that is, $\text{Supp}(T)\subset\{0\}$. If $T$ is not identically zero, this means that $\text{Supp}(T)=\{0\}$.
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0hi, sorry, i don't understand. Can you explain me please why $Supp \delta=\{0\}$, and in some books i found that we haven't equality, just an inclusion, so i'm lost – 2017-02-24
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We can not guarantee that $supp(T)=\{0\}$, because a zero distribution $T=0$ also satisfies this equation, yet $supp(0) = \emptyset$.
In general case, if $f\in C^\infty$ and $T\in D'$ satisfy $fT=0$, then $supp T\subset \{x:\, f(x)=0\}$, this can be easily show by definition, the Julián Aguirre's hint is a good start.
In your case the above result would yield $supp(T)\subset \{0\}$, and on top of that you can write $T$ explicitly: $T=c\delta_0$, $c\in \Bbb C$. The support of $\delta_0$ can be found by definition, too.