Convergence of Fourier series

By David Loeffler (P865) on Wednesday, May 17, 2000 - 11:06 pm :

It follows from Weierstrass' M-test that if the Fourier coefficients of a function form an absolutely convergent series, then the function must be continuous. Is the converse true, i.e. do the coefficients converge absolutely for all continuous functions?

David Loeffler

By Richard Samworth (Rjs57) on Tuesday, May 30, 2000 - 04:51 pm :

David,

The converse is false, but is true if the function is continuously differentiable. This should give you a starting point for finding a counterexample. The reason why it is true if f is continuously differentiable is because in that case, integration by parts shows that

$\hat{f}' (n) = i \times n \times \hat{f} (n)$ for $n \neq 0$

Now consider the sum you're trying to show is finite, ignore the $abs (\hat{f} (0))$ term and use the equation above, the Cauchy-Schwarz inequality on $1 / abs (n)$ and $abs (\hat{f} (n))$ and then Parseval's theorem to give the result.

Please excuse my laziness in not writing this out in full. Exams (e.g. on Fourier Analysis!) looming! Please write back if you'd like more explained, although allow a week or so for delivery!

Best wishes

Richard

By David Loeffler (P865) on Tuesday, June 6, 2000 - 01:45 pm :

Richard,

Sorry, but you've lost me slightly there. I don't really get what your starred equation is saying. (Sorry for taking so long to reply myself but I have had a few A level modules, so I know how you feel!)

David

By Richard Samworth (Rjs57) on Wednesday, June 7, 2000 - 09:39 pm :

David,

The ' denotes differentiation and the ^ denotes the operation of finding the Fourier coefficients. So what this formula gives is an expression for the Fourier coefficients of the derivative of a function in terms of the Fourier coefficients of the original function.

My last exam is tomorrow (Fourier analysis was today!) so I'll write a proper reply soon.
In the meantime please write back to say whether this has helped at all and whether you know what Parseval's theorem, Cauchy-Schwarz etc. are.

Best wishes

Richard

By David Loeffler (P865) on Tuesday, June 13, 2000 - 12:43 pm :

Yes, I see what you're getting at now. I was working with the sine-cosine form of the series rather than the exponential form - in which the same relation almost holds, but not quite.

Anyway, Cauchy-Schwartz:

${(\int_{a}^{b} f (x) g (x) dx)}^{2} \leq (\int_{a}^{b} f (x)^{2} dx) (\int_{a}^{b} g (x)^{2} dx)$

with equality iff $f / g =$ const almost everywhere.

Parseval:

$\sum_{- \infty}^{\infty} c_{n}^{2} = \frac{1}{2 π} \int_{- π}^{π} f (x)^{2} dx$

for all continuously differentiable $f$ .

Are these right?

By Richard Samworth (Rjs57) on Wednesday, June 14, 2000 - 06:47 pm :

David,

You clearly know an awful lot more about Fourier series than I did when I was doing A-level modules! Related to what you said about the sine-cosine and exponential forms of Fourier series, we need to make slight adjustments to your Cauchy-Schwarz and Parseval formulae. I'll try to start pretty much from first principles and hope not to bore you!

We are interested in complex -valued periodic functions, and can think of these as functions on $T = R / (2 π Z$ ). (Don't worry if this doesn't make much sense).

Let $L^{2}$ denote the set of square integrable functions $f : T \to C$ , i.e. the set of functions for which

$\int_{0}^{2 π} | (\mod (f (x {)))}^{2} | dx < \infty$ .

(mod denotes the modulus of a complex number).

We can make the set of $L^{2}$ functions $f : T \to C$ into a complex inner-product space by means of the inner-product

$〈 f, g 〉 = 1 / (2 π) \int_{0}^{2 π} f (x) g (x) * dx$

(Here the $*$ denotes complex conjugation).

Hopefully you can now see that the Cauchy-Schwarz formula for this inner product space needs to be

${(\int_{0}^{2 π} f (x) g (x) * dx)}^{2} \leq (\int_{0}^{2 π} (\mod (f (x {)))}^{2} dx) (\int_{0}^{2 π} (\mod (g (x {)))}^{2} dx)$ .

There is another, related, inner product space which we call $l^{2}$ , namely the set of square-summable complex-valued sequences, that is, sequences $(a_{n})$ with

$\sum_{n = - \infty}^{\infty} (\mod (a_{n} {))}^{2} < \infty$ . See if you can work out the Cauchy-Schwarz formula for this inner-product space - it's this one we're going to need for our proof.

For integrable functions $f : T \to C$ , that is, functions satisfying

$\int_{0}^{2 π} \mod (f (x)) dx < \infty$ ,

we define Fourier coefficients $f^{(n)}$ for all integers $n$ by $f^{(n)} = 1 / (2 π) \int_{0}^{2 π} f (x) e^{- i n x} dx$ .

Parseval's formula is then:

$\sum_{n = - \infty}^{\infty} (\mod (f^{(} n {)))}^{2} = 1 / (2 π) \int_{0}^{2 π} (\mod (f (x {)))}^{2} dx$ ,

for all $L^{2}$ functions $f : T \to C$ .

I don't have any more time right now, but see what you can do with the hints I gave in the previous post as well as this one. PLEASE write back with anything you haven't understood in this message - it has occurred to me that you may not be aware of things like inner-product spaces, $L^{2}$ , $l^{2}$ , etc.!

Good luck, and let me know how you get on. I'll try to write out a full proof when I have time.

Richard

By David Loeffler (P865) on Thursday, June 15, 2000 - 09:45 am :

Is the required Cauchy-Schwarz inequality

${(\sum_{- \infty}^{\infty} \mod (a_{n} b_{n}))}^{2} \leq (\sum_{- \infty}^{\infty} \mod (a_{n}^{2})) (\sum_{- \infty}^{\infty} \mod (b_{n}^{2}))$ ?

I now see what you meant in your original message - applying this version of Cauchy-Schwarz, then Parseval's theorem does show $\sum_{- \infty}^{\infty} \mod (f^{n})$ to be finite.

However, can we extend this in any way to functions which are continuous and piecewise linear? I have tried a number of these and the coefficients seem to tend to zero as $1 / \mod (n)^{2}$ . (For example the ßawtooth" function which is $1 - 2 \times abs (x) / π$ on $[- π, π]$ is $\sum_{- \infty}^{\infty} 2 (1 - (- 1)^{n}) / (n^{2} π^{2}) e^{i n x}$ .) I haven't yet succeeded in finding the counterexample you mentioned in your first message. In fact the "nastiest" continuous function I have found is $\sqrt{x}$ , on $[0, π]$ extended to be odd, which has $f^{n} = O (n^{- 3 / 2})$ (according to Maple, which unlike me can evaluate the rather nasty integrals).

By David Loeffler (P865) on Thursday, June 15, 2000 - 01:33 pm :

Is the general form of the Cauchy-Schwarz inequality

〈 a, b 〉^{2} \leq 〈 a, a 〉 〈 b, b 〉

? As you thought I have only a fairly vague idea of what constitutes an inner product space, although I think I've got a book which gives the definitions somewhere. Incidentally, in the inner product you gave for

L^{2}

, what is there to stop

〈 a, b 〉

being complex and the inequality sign losing its meaning? By the way, sorry for the dodgy formatting in the last message and its abrupt end - I wrote it in rather a hurry!

David.

By David Loeffler (P865) on Friday, June 16, 2000 - 02:41 pm :

Sorry, that Cauchy-Schwarz formula for

I^{2}

is wrong, isn't it?

I guess you would define $〈 a, b 〉 = \sum_{- \infty}^{\infty} a_{n} b_{n} *$ , which now satisfies $〈 a, b 〉 = 〈 b, a 〉 *$ , giving the inequality as

${(\sum_{- \infty}^{\infty} a_{n} b_{n})}^{2} \leq (\sum_{- \infty}^{\infty} \mod (a_{n})^{2}) (\sum_{- \infty}^{\infty} \mod (b_{n})^{2})$ .

By Richard Samworth (Rjs57) on Friday, June 16, 2000 - 02:43 pm :

David,

You're quite right - I made a mistake when I wrote down the inner product for L² - it should have modulus signs around the integral. You were almost right with your general form of Cauchy-Schwarz, except it too needs modulus signs around the inner-product on the left-hand side, as it is in general complex.

Note that < a,a > is real, because it is the square of the norm of a. Have a look at the definitions of an inner-product and norm, and see if you can prove the general Cauchy-Schwarz inequality. Hint: expand < a+tb,a+tb > by linearity as a quadratic in t, use the fact that Re(< a,b > ) < = mod(< a,b > ), and complete the square, noting that the original inner-product is non-negative (it's the square of the norm of a+tb).

With regard to extending the result, piecewise linear continuous functions are differentiable almost everywhere, and the derivative is piecewise constant (so in particular, piecewise continuous). I'm pretty sure it shouldn't be too hard to cover this case, therefore. Have a go and let me know how you get on!
Best wishes

Richard

By Richard Samworth (Rjs57) on Saturday, June 17, 2000 - 11:36 am :

You've got the right definition of the inner product, but from my remark in my last message you'll see that we need modulus signs around the sum on the LHS again!

By David Loeffler (P865) on Sunday, June 18, 2000 - 11:13 pm :

Yes, I have the general Cauchy-Schwarz inequality. I suppose that the conditions for equality drop out quite neatly from this as well since

〈 a + t b, a + t b 〉

can only be zero if

a + t b = 0

for some

t

It seems that any piecewise continuously differentiable function is covered by the original argument but using Bessel's inequality in place of Parseval's theorem. However, I'm not quite sure how restrictive the conditions are for Bessel's inequality to hold.

I did have a totally unrigorous argument that involved differentiating an arbitrary function $k$ times so that $f^{(k)}$ is the first unbounded derivative, and using the fact that $\int_{- π}^{π} \mod (f^{(k)} (x)) dx$ exists, so that $(f^{(k)})^{n}$ is bounded. Then integrating the resulting series $k$ times we would have $f^{n} = O (n^{- k})$ . I have a feeling that this may be a load of total rubbish, but I can't quite see why. Could you suggest a counterexample to this argument, or any way to make it more solid?

David