The distribution solution to $L^{+} u=0$ is u=0, where $L^{+}=-\frac{d}{dx}+x$?

Question

Consider the creation operator $L^{+}=-\frac{d}{dx}+x$. If $u\in L^2(\mathbb{R})$ and is a distribution solution to the equation $L^{+}u=0$, then for any $\phi\in C_0^{\infty}(\mathbb{R})$ we have $\int ((\frac{d}{dx}+x)\phi)u=0 $. But I cannot see why $u=0$.

Answer 1

Let $u$ satisfies

$$\int (\phi'(x) + x\phi(x)) u(x) dx = 0$$

for all compactly supported smooth $\phi$. First we state the following lemma (prove at the end):

Lemma 1 Let $C , D>0$ be constant. Then $$\int (\phi'(x) + x\phi(x)) u(x) dx = 0$$ whenever $\phi$ is smooth, supported in $(A, \infty)$ for some $A$ and $\phi(x) = C e^{-\frac{x^2}{2}}$ when $x >D$.

Now we show:

Lemma 2 $\int g u = 0$ for all $g\in C^\infty_c(\mathbb R)$.

Let $g\in C^\infty_c(\mathbb R)$. Then the function $$\phi(x) := e^{-\frac{x^2}{2}} \left( \int_{-\infty}^x e^{\frac{t^2}{2} }g(t) dt\right)$$ satisfy $\phi'(x) + x\phi (x) = g(x)$ for all $x$. Note that as $g$ is of compact support, $\phi$ satisfies the condition in lemma 1. Thus

$$\int gu = \int (\phi' + x\phi) u = 0.$$

Now we have that $u$ is orthogonal to a dense subset $C^\infty_c(\mathbb R)$ in $L^2(\mathbb R)$, so we must have $u = 0 \in L^2(\mathbb R)$.

It remains to prove lemma 1:

Proof of lemma 1: To see this, let $C' >D$ and $f$ be a smooth function so that $0\le f\le 1$ on $\mathbb R$,

$$f(x) =\begin{cases} 0 & \text{when }x \le C' \\\text{increasing} & \text{when } C'\le x \le C'+1 \\ 1 & \text{when } x\ge C'+1 \end{cases}$$

Then let $g = 1-f$. Now we have

$$\begin{split} \int \left( \frac{d}{dx} + x\right) \phi u &= \int \left( \frac{d}{dx} + x\right) (f+ g)\phi u \\ &= \int \left( \frac{d}{dx} + x\right) f\phi u + \int \left( \frac{d}{dx} + x\right) g\phi u \\ &= \int \left( \frac{d}{dx} + x\right) (f\phi) u \end{split}$$

as $g\phi$ is of compact support. Now as $C' >D$, $\phi (x) = C e^{-\frac{x^2}{2}}$, so

$$ \left( \frac{d}{dx} + x\right) (f\phi) = \frac{df}{dx} \ C e^{-\frac{x^2}{2}}.$$

So by Cauchy-Schwarz inequality,

$$\left| \int \left( \frac{d}{dx} + x\right) \phi u\right| = \left|\int \frac{df}{dx} \ e^{-\frac{x^2}{2}} u\right| \le \sqrt{ \int_{C'}^{C'+1} \left(C\frac{df}{dx}e^{-\frac{x^2}{2}}\right)^2 } \ \|u\|_{L^2}\to 0$$

as we can choose $C' >D$ as large as possible.