Fourier inversion Lemma (Lars Hörmander) 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 $y\in \mathbb R^n$ 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:
$$T \phi(x) = \sum_{j=1}^n(x_j-y_j)T\phi_j(x)=0 \quad \mbox{ if } x=y.$$
(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 $T\phi(x) = c(x) \phi(x)$, and that $c$ is independent of $\phi$. I simply can't see how he arrives at that fact.
 A: 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{(T\phi)(x)}{\phi(x) }
\end{equation}
Then naturally we have that $T\phi = c\phi$, 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 $f\in \mathcal S(\mathbb R^n)$ 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)(T\psi)(x)-\psi(y)(T\phi)(x) \quad \Rightarrow\\
  (Tf)(y)&=\phi(y)(T\psi)(y)-\psi(y)(T\phi)(y).
\end{align}
Hence for the fixed point $y$ we have that $0=\phi(y)(T\psi)(y)-\psi(y)(T\phi)(y)$, which imply that:
\begin{equation}
 (T\psi)(y)=\psi(y) \frac{(T\phi)(y)}{\phi(y)} = c(y)\psi(y).
\end{equation}
Since both $\psi$ and $y$ where arbitrarily chosen we have the desired result.
