I encountered an exercise in Lars Hörmander's book "The Analysis of Linear Partial Differential Operators I". The exercise asks to prove the differential operator $L$ defined below is surjective on $\mathcal{S}(\mathbb{R})$, or, equivalently, to solve the following differential equation: $$ Lf = xf(x) + f^\prime(x) = g, \forall g \in \mathcal{S}(\mathbb{R}), $$ where $\mathcal{S}(\mathbb{R})$ denotes the Schwartz space.
The original statement of this is about multivariate functions, but I do believe the essence lays in the one variable case already. What we can observe is that the differential operator $L$ commutes with the Fourier transform $\mathcal{F}$ up to $i$ i.e. $\mathcal{F}(Lf) = iL(\mathcal{F}(f))$. But I don't know how to do the next.
Any comments or solutions will be very much appreciated!
Showing the existence of a solution.
Write
$$\phi'(x)=F(x,\phi(x))\tag 1$$ Where $$F(p,q)=g(p)-pq$$ With some initial condition, say $\phi(0)=1$. If $g$ is Schwarz, then it is continuous, and therefore $F$ is continuous in its first argument. Because $F$ is simply linear in its second argument, it is therefore Lipschitz continuous in its second argument and therefore by the Picard–Lindelöf theorem there exists an $\epsilon > 0$ s.t a solution to $(1)$ exists in $(-\epsilon,\epsilon)$.
Following the formalism established in this section, the optimal $\epsilon$ is found to be $$\epsilon_{\text{opt}}=\min\left(a,\frac{b}{M}\right)$$ Here $M=\Vert F\Vert_\infty=\sup |F|$ and $a,b$ represent the domain of values for which $F$ is defined in its first and second arguments respectively, i.e $$\operatorname{dom}F(\cdot,q)=(0-a,0+a) \\ \operatorname{dom}F(p,\cdot)=(0-b,0+b)$$
Since $g$ is Schwarz, it is defined on all of $\mathbb R$, and hence $a=\infty$. Since $F$ is linear in its second argument clearly $b=\infty$ as well and thus $\epsilon_{\text{opt}}=\infty$, i.e the solution for $\phi$ exists on all of $\mathbb R$.
So far, we have found that
Showing that the solution is Schwarz.
Again return to $(1)$.
$$\phi'(x)=g(x)-x~\phi(x)$$ Because $g$ is Schwarz, we know that $g(x)\to 0$ rapidly as $x\to \pm \infty$. This means that for large $x$, we know that $$\phi'(x)\approxeq -x~\phi(x) \\ \text{as}~x\to\pm\infty$$ Which, along with the initial condition, implies $$\phi(x)\approxeq \exp(-x^2/2) \\ \text{as}~x\to\pm\infty$$
So defining $\Vert \phi\Vert_{m,n}=\sup_{x\in\mathbb R} |x^m \phi^{(n)}(x)|$ then we can at least be confident that $\Vert\phi\Vert_{m,0}<\infty$.
Now, we can have a look at the $n=1$ case, $$x^m\phi'(x)=x^mg(x)-x^{m+1}\phi(x) \\ \Vert\phi\Vert_{m,1}=\sup_{x\in\mathbb R}|x^m\phi'(x)|=\sup_{x\in\mathbb R}|x^m g(x)-x^{m+1}\phi(x)|$$ Clearly, by the triangle inequality $$|x^m g(x)-x^{m+1}\phi(x)|\leq |x^m g(x)|+|x^{m+1}\phi(x)|$$ Hence $$\Vert\phi\Vert_{m,1}\leq \sup_{x\in\mathbb R}|x^mg(x)|+\sup_{x\in\mathbb R}|x^{m+1}\phi(x)|=\Vert g\Vert_{m,0}+\Vert\phi\Vert_{m+1,0}<\infty$$ Since $g$ is Schwarz and by the previously established result for the $n=1$ case.
How do we continue this for $n=2,3,4,\dots$ etc? Well, taking derivatives of $(1)$, see that $$x^m\phi^{(n+1)}(x)=x^m g^{(n)}(x)-x^m\frac{\mathrm d^n}{\mathrm dx^n}(x~\phi(x))\tag{2}$$
I'll let you check for yourself (induction) that $$\frac{\mathrm d^n}{\mathrm dx^n}(x~\phi(x))=n\phi^{(n-1)}(x)+x~\phi^{(n)}(x)$$
Therefore $(2)$ is just $$x^m\phi^{(n+1)}(x)=x^m g^{(n)}(x)-nx^m\phi^{(n-1)}(x)-x^{m+1}\phi^{(n)}(x)$$
Hence
$$\Vert\phi\Vert_{m,n+1}=\sup_{x\in\mathbb R}|x^m\phi^{(n+1)}(x)|=\sup_{x\in\mathbb R}|x^m g^{(n)}(x)-nx^m\phi^{(n-1)}(x)-x^{m+1}\phi^{(n)}(x)| \\ \leq \sup_{x\in\mathbb R}|x^m g^{(n)}(x)|+\sup_{x\in\mathbb R}|nx^m\phi^{(n-1)}(x)|+\sup_{x\in\mathbb R}|x^{m+1}\phi^{(n)}(x)| \\ \implies \Vert\phi\Vert_{m,n+1}\leq \Vert g\Vert_{m,n}+n\Vert\phi \Vert_{m,n-1}+\Vert\phi\Vert_{m+1,n}$$
Therefore the rest of the needed constants $\Vert\phi\Vert_{m,2},\Vert\phi\Vert_{m,3},\dots$ are all finite by induction, and hence $\phi$ is Schwarz.
With these two points is established,
We have shown that $L$ is surjective on $\mathcal S(\mathbb R)$.
QED. $\blacksquare$