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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.

Martin
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  • Here is a line of thought that I've done: If $\phi(y)=0$ then we can prove (from the argument above) that $T\phi(y)=0$, but since $T \colon \mathcal S (\mathbb R^n) \to \mathcal S (\mathbb R^n)$ then we have that $T^m\phi(y) =0$ for all $m \in \mathbb N$. I still have no idea if this is useful or not. – Martin Apr 29 '15 at 05:54
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    Just define $c(x)$ by this equation. To see that it's independent of $\varphi$, observe that $(T\varphi)(x)$ only depends on $\varphi(x)$, not on the function as a whole (consider the difference of two functions that take the same value at $x$ to see this). So $c(t)\varphi(t)$ for a given $t$ can be obtained by applying $T$ to $\varphi(t)f(x-t)$, where $f$ is a fixed function with $f(0)=1$. –  May 16 '15 at 04:32
  • this is just a muse, but the $c(i)$ might be a concept parallel to this one: If $v_1,..,v_n$ is a basis, and $\alpha_i,..,\alpha(n)$ a set of scalars, ${\alpha_i.v_i}$ is also a basis – JMP May 22 '15 at 04:19
  • This is a comment on the other post, the one you just deleted - can't comment there. You deleted it too fast! If you undelete it (and let me know you've done so) you'll get a big chuckle out of the answer I'd like to post - it's much simpler than we thought, no Baire category needed... – David C. Ullrich Jul 16 '16 at 13:31
  • It's undeleted :o) – Martin Jul 17 '16 at 06:05

1 Answers1

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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.

Martin
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