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I am seeing the Schwartz Space for the first time today and I have trouble understanding the following argument: Given that $f \in S({\mathbb{R}})$ with $f(x_0) = 0$ Taylor's theorem tells us that we can find a function $g \in C^\infty(\mathbb{R})$ such that \begin{equation} f(x) = g(x)(x - x_0) \end{equation} Hence, for $x \neq x_0$ we can write $g(x)$ as \begin{equation} g(x) = (x - x_0)^{-1} f(x) \end{equation} According to my notes, from this it should be clear that $g \in S(\mathbb{R})$, but how can I see this immediately ? In order to derive this I think would need to direvtly go via the definition, i.e. show for each $k,m = 0,1,2, \dots$ there exist constants $c_{k,m}$ such that \begin{equation} \sup_x |x^k g^{(m)}(x)| \leq c_{k,m} \end{equation} or is there a shorter way to infer this ? By "shorter" I mean for example a statement that somehow the constants that bound $f$ are also bounds on $g$, probably with some amendment.

Edit: Does it maybe have to do with the fact that $f$, being smooth, has a Tayler expansion and so \begin{equation} g(x) = \sum^\infty_{n = 1} \frac{f^{(n)}(x_0)}{n!}(x - x_0)^{n-1} \quad (\text{note } f(x_0)) = 0 \end{equation}

Thanks for your help!

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up vote 3 down vote accepted

The function $f(x)$ does not have to have a Taylor expansion: smoothness is a much weaker condition (for real analysis) than having a series expansion. In fact the existence of that function $g(x)$ is a subtle issue which can be settled without power series expansions (which may not exist) by a beautiful trick with the fundamental theorem of calculus.

To keep things simple, let's take $x_0 = 0$, so $f(0) = 0$. We want to show $f(x) = xg(x)$, where $g(x)$ is a Schwartz function. Well, $$ f(x) = f(x) - f(0) = \int_0^x f'(t)\,dt = x\int_0^1 f'(xu)\,du, $$ where in the second equation we used the Fundamental Theorem of Calculus and in the third equation we made a multiplicative change of variables $t = xu$.

Staring at the above equations, define $g(x) = \int_0^1 f'(xu)\,du$. This makes sense for all $x$ and we have shown $f(x) = xg(x)$ for all $x$. Now it's up to you to show that if $f(x)$ is infinitely differentiable then $g(x)$ is infinitely differentiable and if $f(x)$ is in the Schwartz space then $g(x)$ is. You just need to have an appropriate theorem on differentiating under the integral sign.

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great answer! I'll make sure I revise the theorem I need to differentiate under the integral sign, I remember this being mentioned at the end of my course on analysis .. anyways, many thanks!! – harlekin Jan 31 '12 at 8:49

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