# Fourier transform smoothness

1. May 29, 2010

### mnb96

Hello,
I read somewhere that if a function f decays rapidly (e.g. $\lim_{x \to \infty}f(x)=0 )$, then its Fourier transform F is smooth.
How can I prove this? (Reference to some sources are welcome too).
Thanks.

2. May 29, 2010

### Jerbearrrrrr

I think f(x) has to tend to zero for the fourier transform to even exist.
Just did some musing. If f and all its derivatives decay faster than any polynomial (eg, exponentially fast), F should be smooth. I'm sure there are less strict conditions though.

Sometimes the FT is studied as an isomorphism from the Schwartz space to itself (functions from ___ to ___ whose derivatives of all orders and types decay exponentially fast. Sometimes it's expressed as having a certain finite norm)

3. May 29, 2010

### mnb96

do you have a reference to a proof for this statement?
It doesn´t look like a trivial result, at least to me.

4. May 29, 2010

### Jerbearrrrrr

My lecture notes lol. You need to use the fact that FT(f ') = k*FT(f), and the inverse/dual (not sure precisely what to call it) of this fact.

From my PDEs course last term. It's actually like the first thing that was proven after defining the FT:

Defn (Fourier Transform)
for $$f \in L^1(R^n;C) \ \ \ \ \ \ \ \ \hat{f} (\mu) := \int _{R^n} f(x) e^{-ix.\mu} dx$$
$$(\mu \in R^n, \hat{f} : R^n \to C)$$

Lemma
Let $$f \in S(R^n)$$. Then $$\hat{f} \in S(R^n)$$.
Proof
For any multi-indices $$k_1, k_2$$
$$sup_\mu | \mu^{k_1} \nabla ^{k_2} \hat{f} (\mu)| \leq sup |F(\nabla _x ^{k_1} (x^{k_2} f))(\mu)|\leq | | \nabla _x ^{k_1}(x^{k_2} f) | | _{L^1}$$

I've sloppily used F(f) to be the fourier transform of f that changes x to mu. The ||.|| thing is the L1 norm which is just "integrate me over all space".
It's presented as a one liner but I guess there are quite a few ideas involved.

Oh, and S is
$$S:= \{f \in C^{\infty} (R) : lim_{x \to \infty} \frac{|\nabla ^k f(x) |}{|x| ^m} \to 0 \ \ \forall k,m\}$$
So thingies in S are definitely smooth and decaying.

Last edited: May 29, 2010
5. May 31, 2010

### mnb96

Last edited by a moderator: Apr 25, 2017
6. May 31, 2010

### Jerbearrrrrr

Oh. Yes, it is clear.
Compact support means (well it doesn't, but it does) the function is non-zero only locally (in some closed set).

ie it looks like a bump.

7. Jun 1, 2010

### mnb96

I tried to prove it this way (by induction on the power of $|x|^m$).
The cases when $m\geq 0$, are trivial so we assume m<0, and f is smooth with $\lim_{x \to \infty}f(x)=0$.
Let k=-m

Base (k=1):

$$\lim_{x\to +\infty} xf^{(n)}}(x) = \lim_{x\to +\infty} \frac{x}{[f^{(n)}(x)]^{-1}} = \lim_{x \to \infty}\frac{f^{(n+1)}}{-[f^{(n)}(x)]^{-2}} = 0$$

Induction:

$$\lim_{x\to +\infty} x^{i+1}f^{(n)}}(x) = \lim_{x\to +\infty} \frac{x}{[x^{i}f^{(n)}(x)]^{-1}}= \lim_{x\to +\infty} \frac{1}{D[(x^{i}f^{(n)}(x))^{-1}]} = 0$$

The last step is due to the fact that the denominator is the product of two smooth functions (which by induction hypothesis tends to 0, and so all its derivatives).

If this is correct, it looks more clear to me. Perhaps I don't clearly understand the definition of function with compact support (at least I don't know how to use that definition).

Last edited: Jun 1, 2010
8. Jun 1, 2010

### Jerbearrrrrr

[PLAIN]http://img707.imageshack.us/img707/2655/compsupp.png [Broken]
This is a smooth function with compact support. (Assume it's smooth anyway)

It's only non-zero locally.
So towards infinity...of course it decays fast, because it IS zero.

Last edited by a moderator: May 4, 2017