Give an example to show that not all smooth rapidly decreasing functions are Schwartz fiunctions.

Question

Give an example to show that not all smooth rapidly decreasing functions are Schwartz functions.

hint is provided: Try to come up with a function

$$f(x)=\sum_k\frac 1 {2^k} f_k(x) \text{ where each of } f_k\in C_c^\infty(\mathbb{R})$$

i am wondering what kind of $f_k$ can i take here? any suggestion

Answer 1

You could let each $f_k$ be a "bump" function: a nonnegative smooth function that vanishes off a ball.

One way to go about this is to start with a function $g$ that satisfies the following:

• $g \in C^\infty(\mathbb R)$,
• $0 \le g(x) \le 1$,
• $|x| \ge 1$ implies $g(x) = 0$,
• $g(0) = 1$.

There are specific examples of such functions, but we can take the existence of such a $g$ for granted.

Now let $f_k(x) = f(2^k(x-k))$. You should be able to verify that $f$ defined as in the question is rapidly decreasing. The graph of $f$ consists of a bunch of nonoverlapping bumps whose heights decrease to $0$. On the other hand, $f'(x) \not\to 0$ as $x \to \infty$ because of the cancellation of the $2^k$'s.