Solve $xu'+u=0$ as distribution

If $u \in \mathcal{D}'(\mathbb{R})$ where $\mathcal{D}'(\mathbb{R})$ is the space of all tempered distributions defined on $C_c^{\infty}(\mathbb{R})$ and $xu'+u=0$. Show that $u = A \mathbf{p.v.}\frac{1}{x} + B\delta$, where $$\mathbf{p.v.}\frac{1}{x}(f(x)) = \lim_{\epsilon \to 0^+} \int_{|x|\ge\epsilon} \frac{f(x)}{x} \mathbf{d}x, \quad f \in \mathcal{D}(\mathbb{R})$$

I have already shown that $\mathbf{p.v.}\frac{1}{x}$ is a tempered distribution. For $f \in \mathcal{D}(\mathbb{R})$, we have \begin{align} 0 &= \langle xu', f \rangle + \langle u, f \rangle \\ &= \langle u', xf \rangle + \langle u, f \rangle \\ &= -\langle u, f+ xf' \rangle + \langle u, f \rangle \\ &= -\langle u, xf' \rangle \\ &= -\langle xu, f'\rangle \\ &= \langle (xu)', f\rangle \end{align} Since this holds for any $f$, it must be true that $xu=0$.

But I don't know where to go next.

• No, it doesn't follow that $xu=0$; what follows from the last line is that $xu$ is a constant. – David C. Ullrich Dec 5 '17 at 15:22
• You only have $\langle xu, g \rangle = 0$ on a subspace of $\mathcal D(\mathbb R),$ namely for those $g \in \mathcal D(\mathbb R)$ which can be written as a derivative of some $f \in \mathcal D(\mathbb R).$ – md2perpe Dec 5 '17 at 15:30
• @DavidC.Ullrich If $xu=c$ and $f \in \mathcal{D}$, then $f' \in \mathcal{D}$ and so $c = \langle xu, f' \rangle = 0$. Am I wrong? – Alan Zhang Dec 5 '17 at 17:07
• @DavidC.Ullrich What does it mean for $xu$ to be constant? – Alan Zhang Dec 5 '17 at 17:09
• @AlanZhang It means there exists $c\in\mathbb C$ such that $xu=c$, which in turn means $<xu,\phi>=c\int\phi$. What else might it mean? (This is equivalent to $(xu)'=0$.) – David C. Ullrich Dec 5 '17 at 17:26

$\newcommand{\tf}{\mathcal D}$

Hint: If $u\in\tf'$ and $u'=0$ then $u$ is constant: There exists $c\in\mathbb C$ such that $u=c$.

Regarding what that means and then why it's true: First, saying $u'=0$ means $\langle u',\phi \rangle=0$ for all $\phi\in\tf$, which by the definition of $u'$ means $$\langle u,\phi' \rangle=0\quad(\phi\in\tf).$$

In general, if $u\in\tf'$ and $f$ is a locally integrable function then saying $u=f$ means $$\langle u,\phi \rangle=\int f\phi\quad(\phi\in\tf).$$

SO what's being asserted is this:

If $u\in\tf'$ and $\langle u,\phi' \rangle=0$ for all $\phi\in\tf'$ then there exists $c\in\mathbb C$ such that $\langle u,\phi \rangle= c\int\phi$ for all $\phi\in\tf'$.

Proof:

Exercise Suppose $\phi\in\tf$. There exists $\psi\in\tf$ with $\phi=\psi'$ if and only if $\int\phi=0$.

Now suppose $u'=0$. Choose $\phi_0\in\tf$ with $\int\phi_0=1$ and let $$c=\langle u,\phi_0 \rangle.$$ Suppose $\phi\in\tf$. Let $\alpha=\int\phi$. The exercise show that there exists $\psi\in\tf$ with $$\phi-\alpha\phi_0=\psi'.$$Hence $$(u,\phi-\alpha\phi_0 \rangle=0,$$or $$\langle u,\phi \rangle=c\int\phi.$$

Now to finish your problem: You have $(xu)'=0$, so $xu=c$. If you let $v=u-c\frac1x$ it follows that $xv=0$, and now an argument much like what's above shows that $v$ is a multiple of $\delta$. (Hint for that last bit: If $\phi\in\tf$ and $\phi(0)=0$ then there exists $\psi\in\tf$ with $\phi=x\psi$.)