Class of infinitely differentiable and bounded complex value functions

Question

The following questions is from Real and Complex Analysis editted by Walter Rudin. It is the 5th exercise in Chapter 9 (the one about Fourier Transform).

Let $m$ be the Lesbegue Measure on the real line and define $\int_{-\infty}^{\infty}f(x)dm(x) = \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}f(x)d(x)$Let $S$ be all functions $f: \mathbb{R} \rightarrow \mathbb{C}$ that are infinitely differentiable and for any $m, n \in \mathbb{N}$, there exists a real number $A_{mn}(f)$ such that $sup|x^nf^{(m)}(x)| \leq A_{mn}(f)$. Now prove that the fourier transform mapping maps $S$ onto $S$.

According to Theorem 9.2, point (2) inside the book, when $f \in L^1$, $(\hat{f})^{(1)}(t)$ = $-i \int^{\infty}_{-\infty} xf(x)e^{-ixt}dm(x)$. However, in this question, I do not know whether or not $f$ is in $L^1$. If it is, how to find an upper bound when taking integration along the real line?

Answer 1

The point of $A_{mn}(f)$ is control of tail behavior: these functions decrease faster than every polynomial. So we should use this condition to estimate $\int |f|$ near infinity:

$$ \int_1^\infty |f|\ dm = \int_1^\infty |x^{-2}||x^2f|\ dm \leq \sup |x^2f| \int_1^\infty |x^{-2}|\ dm \leq A_{21}\int_1^\infty x^{-2}\ dm < \infty$$

Use other means to control $\int|f|$ on compact intervals and you're set.

Answer 2

Sketch. The part after "Let $m$ be the Lesbegue Measure on the real line and define" seems to be confused and not relevant so I'll ignore.

I do not know whether or not $f$ is in $L^1$.

Well, note that $|f(x)| \le A_{00}$ and $|f(x)|\le \frac{A_{02}}{|x|^2}$. In particular, $f\in L^\infty$ and $|f|\le C |x|^{-s}$ for $s$ larger than the dimension (i.e. $s>1$). This implies $f\in L^1$.

how to find an upper bound

You want to bound $\mathscr F f(\xi)=\int_\mathbb R f(x) e^{ix\xi} dx$. Or something similar; maybe I'm missing a minus sign or a multiple of $\pi$ in a few places. Lets pretend I'm right, because it doesn't matter. More specifically, you want to check that for any $m,n$ there exists $A_{mn}(\mathscr F f)$ such that $$\sup_{\xi\in\mathbb R} |\xi|^n|(\mathscr F f)^{(m)}(\xi)|<A_{mn}(\mathscr F f).$$ You should be able to use the following facts, and similar manipulations as above, to conclude:

1. $ |\xi \mathscr F g| = |\mathscr F(g')|$
2. $ |(\mathscr Fg)'| = |\mathscr F(x g)|$
3. $ \int |fg| \le \|f\|_{L^\infty} \|g\|_{L^1}$