Do Fourier Coefficients Decay with Increasing n?

[s_j_sawyer]

Homework Statement

Let f be a C1 function on [-pi,pi]. Prove the Fourier coefficients of f satisfy

|an| <= K/n and |bn| <= L/n n=1,2,...

Homework Equations

an = 1/pi * int[-pi..pi] (f(x)*cos(nx)) dx

bn = 1/pi * int[-pi..pi] (f(x)*sin(nx)) dx

Sorry if my form is slightly unpleasing to the eye, but I'm sure if you're reading my post you probably know what I'm talking about.

The Attempt at a Solution

|an| = | 1/pi * int[-pi..pi] (f(x)*cos(nx)) dx |

<= 1/pi * int[-pi..pi] | (f(x)*cos(nx)) | dx

and that's as far as I could get. I thought maybe I could show that

int[-pi..pi]( | cosnx | )dx <= (1/n)*int[-pi..pi]( |cosx| ) dx

but that turned out to be false.

Any ideas?

 

[lanedance]

not there, but some ideas that are hopefully helpful
so I = $\left[ -\pi, \pi \right]$ is a closed interval

as the function is $C^1$ both f & f' are continuous on I and so attain a maximum on I

$$f = a_0 + \sum_{n=1} (a_n cos(nx) + b_n sin(nx))$$

$$f' = \sum_{n=1} (-n.a_n sin(nx) + n .b_n cos(nx))$$

as f and f' are bounded, then applying Parseval's theorem, for some positive number M
$$\int f'^2 = M$$
$$= \pi \left( \sum_{n=1} (n.a_n)^2 + (n .b_n)^2 \right)$$

 

[s_j_sawyer]

Ok I followed everything you said but I still don't see how this relates to what I'm trying to show. The fact that there is a /n necessary is really confusing me.

i.e. |an| <= K/n and not just KEdit:

Ok I may have gotten it.

I think the solution is |an| <= |a1|/n and |bn| <= |b1|/n

There's too much to write out how I got this but does this seem correct?

 

[vela]

Ok I followed everything you said but I still don't see how this relates to what I'm trying to show. The fact that there is a /n necessary is really confusing me.

i.e. |an| <= K/n and not just K

The 1/n factor is the reason lanedance looked at the series for f' rather than f. Look at the series from Parseval's identity. What can you say about the sign of each term?

Edit:

Ok I may have gotten it.

I think the solution is |an| <= |a1|/n and |bn| <= |b1|/n

There's too much to write out how I got this but does this seem correct?

No, this isn't correct. f(x)=sin(2x) is a simple counterexample since it has a1=0 and a2=1.