Xrfgtjtk

I always like to have more than one proof for the same theorem. The other day I was browsing through my copy of Lars Hörmander's book on PDE (volume 1).
When proving the fourier inversion formula (on $mathcal{S}(mathbb{R}^n)$) he makes use of the following lemma:

If $T colon mathcal{S}(mathbb R^n) to mathcal S (mathbb R^n)$ is a linear map such that:
$$TD_j phi = D_j T phi$$ and
$$Tx_j phi = x_j T phi$$ for all $j in { 1 , ldots n}$ and $phi in mathcal S (mathbb R^n)$. Then $T phi = c phi$, for some constant $c$.

In the proof of this lemma he shows that if $phi (y)=0$, for some $yin mathbb R^n$ then $phi$ can be written in the following form:
$$phi(x) = sum_{j=1}^n {(x_j -y_j)phi_j(x)}quad mbox{with } phi_j in mathcal S (mathbb R^n)$$
(this is not the problem - as he gives a good hint as to how to construct the $phi_j$'s).

He goes on showing that:
$$T phi(x) = sum_{j=1}^n(x_j-y_j)Tphi_j(x)=0 quad mbox{ if } x=y.$$
(this is also really simple - but now comes the tricky part).

He goes on to conclude that there exist some function $c(x)$ such that $Tphi(x) = c(x) phi(x)$, and that $c$ is independent of $phi$. I simply can't see how he arrives at that fact.

So here's an answer to my own question. Just in case it would be of interest for someone else.

Firstly take $phi in mathcal S(mathbb R^n)$ such that $phi >0$. Now we may define $c$ by:
begin{equation}
c(x) = frac{(Tphi)(x)}{phi(x) }
end{equation}
Then naturally we have that $Tphi = cphi$, for this particular choice of $phi$.

Now take an arbitrary $psi in mathcal S(mathbb R^n)$, and some fixed point $y in mathbb R^n$. Define a function $fin mathcal S(mathbb R^n)$ by:
begin{equation}
f(x) = psi(x)phi(y)-phi(x)psi(y).
end{equation}
Note that $f(y)=0$. Thus we know that $(Tf)(y)=0$. On the other hand we have that:
begin{align}
(Tf)(x)&=phi(y)(Tpsi)(x)-psi(y)(Tphi)(x) quad Rightarrow\
(Tf)(y)&=phi(y)(Tpsi)(y)-psi(y)(Tphi)(y).
end{align}
Hence for the fixed point $y$ we have that $0=phi(y)(Tpsi)(y)-psi(y)(Tphi)(y)$, which imply that:
begin{equation}
(Tpsi)(y)=psi(y) frac{(Tphi)(y)}{phi(y)} = c(y)psi(y).
end{equation}
Since both $psi$ and $y$ where arbitrarily chosen we have the desired result.