# Showing a function is Schwartz

This question is related to this post, for which I received a really good answer that gave a beautifull solution, yet I am still trying to understand one thing on the side that is not covered by the answer, which is the following claim:

Suppose we take a Schwartz function $f \in S(\mathbb{R})$ that satisfies $f(x_0) = 0$. Then, we have

$$g(x) = (x - x_0)^{-1}f(x) \quad \in S(\mathbb{R}) \quad (x \neq x_0)$$

I am struggeling to come up with an idea to show this without reverting to the plain definition of a Schwartz function and trying to find the bounds on $g$. But is there another , more immediate answer? From my notes it sounds like there is, but there are no immediate steps.

The answer in the post that I linked this one to actually constructs g specifically, however in my notes the claim is made without any reference to a construction so I am guessing that I miss something here with regards to the properties of g that are imposed on it simply by the equation above... Thanks so much for your help!

-

You want to define $g(x_0) = f'(x_0)$ as well.
You might note that $g(x) = \int_0^1 f'(x_0 + s(x-x_0)) \ ds$; this makes it easy to show that $g$ is $C^\infty$ and satisfies the appropriate bounds.
 I am afraid I need to ask - how do I show $g$ is smooth ? Sorry for this dumb question! – harlekin Jan 31 '12 at 22:19 $g'(x) = \int_0^1 s f''(x_0 + s(x - x_0))\ ds$. To see this is correct, look at $\int_{x_0}^x \int_0^1 s f''(x_0 + s (t -x_0))\ ds \ dt$ and interchange the order of integration. Repeat... – Robert Israel Jan 31 '12 at 23:17 okay, thanks a lot !! – harlekin Jan 31 '12 at 23:39