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Is it possible to have a distribution $u\in \mathcal{D}'(\mathbb{R})$ such that the restriction of $u$ to $(0,\infty)$ is $\frac{1}{x}$ and the restriction of $u$ to $(-\infty,0)$ is $0$?

It seems to me that the answer should be 'no', but I couldn't figure how to prove it.

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1 Answer 1

It is possible: define $u\in\mathcal D'(\mathbb R)$ by $u(\phi)=\int_0^\infty \frac 1 x (\phi(x)-\phi(0)e^{-x}) dx$.

Note that for any test function $\phi$, the function $x\mapsto \phi(x)-\phi(0)e^{-x}$ is smooth and sends zero to zero (and has compact support), which implies that $(\phi(x)-\phi(0)e^{-x})/x$ can be integrated on $(0,\infty)$. (Thanks Julián Aguirre for pointing out the problem with just integrating $\phi(x)/x$.)

$u$ is clearly linear. To show that $u$ is a distribution, consider a sequence $\phi_n\to 0$ in $\mathcal D(\mathbb R)$. The support of the functions $\phi_n$ is contained in some interval $[-K,K]$ (and we can assume $K\geq 1$). By integration by parts we have $$u(\phi_n)=\lim_{\epsilon\to 0^+}\int_\epsilon^K \frac 1 x (\phi_n(x)-\phi_n(0)e^{-x}) dx =-\lim_{\epsilon\to 0^+}\int_\epsilon^K \log(x) (\phi_n'(x)+\phi_n(0)e^{-x}) dx$$

Note $\int_0^K|\log x|dx\leq K\log K+1$. Thus $$|u(\phi_n)|\leq (K\log K+1) \max_{0\leq x\leq K}(|\phi_n'(x)|+|\phi_n(0)|)\to 0$$

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Note that $1/x$ is not locally integrable at $x=0$, so that $\phi(x)/x$ is not integrable if $\phi(0)\ne0$. –  Julián Aguirre Mar 13 '13 at 17:28
    
@JuliánAguirre: thanks for pointing out the problem. –  Colin McQuillan Mar 13 '13 at 17:36
    
Thanks, but I still have one silly question: why is $u$ a distribution? i.e. how can I find an appropriate bound to $|u(\phi)|$ given a compact set $K\subseteq \mathbb{R}$? –  Dist Mar 13 '13 at 21:11
    
@Dist: I've added an argument bounding $|u(\phi)|$. –  Colin McQuillan Mar 13 '13 at 23:40
    
I'm probably missing something silly again, but why does $u(\phi)=\int_0^\infty \frac{\phi(x)}{x}$ for $\phi$ with compact support? –  Dist Mar 14 '13 at 15:10
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