Help with understanding a proof in Fourier Analysis

I have a stack of lecture notes that I am currently going through to teach myself a little bit about Fourier Analysis. Now I struggle with the following Lemma, which is needed to talk about the Inverse Fourier Transform in $S(\mathbb{R})$:

Let $T : S(\mathbb{R}) \to S(\mathbb{R})$ be a linear map that satisfies $T(xf) = xT(f)$ and $(Tf') = (Tf)'$ for all functions $f \in S(\mathbb{R})$. Then $Tf(x) = cf(x)$ where $c$ is some constant.

Here is the Proof: (I have inserted my questions in bold as the proof proceeds)

Let $f \in S(\mathbb{R})$ and $f(x_0) = 0$. Then by Taylor's formula, $$f(x) = (x - x_0)g(x) \quad \text{with some } g \in C^\infty(\mathbb{R}).$$

We have $g(x) = (x - x_0)^{-1}f(x)$ for $x \neq x_0$, so $g \in S(\mathbb{R})$.

Q1: How do I know $g \in S(\mathbb{R})$ just from this equation ? I obtained very nice answers in previous posts where I had isolated this problem. However, all of them construct g by adding more information than is given here, i.e. they give g as an integral. Do I need this or is it valid to get to the conclusion as above? To me its seems the author of derives his conclusion from the fact that $g = (x - x_0)^{-1}f(x)$ where $f$ is Schwartz .. is that right ? That is, does this hold in general?

Now we obtain $$Tf(x_0) = T((x - x_0)g(x)\mid_{x = x_0} = (x - x_0)Tg(x)\mid_{x = x_0} = 0$$

Since $T$ is a linear map, this implies that

$Tf_1(x_0) = Tf_2(x_0)$ if $f_1(x_0) = f_2(x_0)$.

Let $f_0 \in S(\mathbb{R})$ be a function such that $f_0(x_0) = 1$ and let $z_0 = Tf(x_0)$.

Since $T$ is a linear map, for an arbitrary function $f \in S(\mathbb{R})$ we have $Tf(x_0) = z_0f(x_0)$.

Q2: Where is the $f_0$ gone ? Here I guess is some typo, the above argument doesn't make sense to me but I don't know how to amend it so that it is correct, any hint as to where the $f_0$ is mixed up with the function $f$ would be helpful! (I realize this is a dumb question, sorry for this, I am really lost in this proof.)

Let $\phi$ be the function defined by $\phi(x_0) = z_0$ for all $x_0 \in \mathbb{R}$. Then by the preceding we have $$Tf(x) = \phi(x)f(x)$$ Since $Tf(x) \in S(\mathbb{R})$, the function $\phi$ is differentiable. Now $$\phi f' = (Tf') = (Tf)' = (\phi f)' = \phi f' + \phi 'f, \quad \forall f \in S(\mathbb{R})$$ This implies that $\phi \equiv 0$; i.e. $\phi$ is identically equal to some constant. $\\q.e.d$

Q3: Where do I need the property of rapid decay, i.e. does this proof work for such a linear map defined on other spaces (e.g. $L_2(\mathbb{R})$ as well ?

By $S$, do you mean the Schwartz class? –  robjohn Feb 1 '12 at 0:29
For the first question, the fact that $g$ is Schwartz follows from writing $g(x)=\int_0^1 f'(tx+(1-t)x_0)dt$. –  Jeff Feb 1 '12 at 3:28