Coefficients from Stone Weierstrass versus Fourier Transform - MathOverflow

Coefficients from Stone Weierstrass versus Fourier Transform

2010-08-24T13:13:44Z

Usually one shows the density of the functions $\sin(kx)$ in $L^2([0,1])$ using the Fourier transform. This in fact comes from the Stone-Weirstrass theorem however and then uses the density of continuous functions in $L^2([0,1])$.

However, the Stone Weirstrass theorem can be used to show, for example, that the functions $e^{ikx}$ are dense in $C([0,1])$ and hence dense in $L^1([0,1])$ as well. So we obtain (not-necessarily-unique) coefficients $c_k$ such that $f_k(x) =c_ke^{ikx}$ converge to any given $f \in L^1([01])$. How should I think about these coefficients? How do they relate to the Fourier series of $f$ (with basis $e^{ikx})$?

Answer by Otis Chodosh for Coefficients from Stone Weierstrass versus Fourier Transform

2010-08-24T18:43:25Z

Re-reading your question, I think that I see what you are asking.

Per @Andrea Ferretti's comments, you have to be careful to distinguish between ${e^{inx}}$ and $span \ {e^{inx}}$. You certainly are interested the latter. Sorry if my comments were sloppy and confusing above.

So, I think that the it goes like this:

From some corollary of Stone-Weirstrauss you can show that $span \ {e^{inx}}$ is dense in $C(\mathbb{S}^1)$ with the supremum norm. Because we know that $C(\mathbb{S}^1)\hookrightarrow L^2([0,1])$ has its image a dense subset of $L^2([0,1])$ and we know that if $f_n \to f$ in the supremum topology on $C(\mathbb{S}^1)$, then the images also converge in $L^2([0,1])$.

Thus, by this reasoning, for $f\in L^2([0,1])$ we can find $f_n \in span \ {e^{inx}}$ such that $f_n = L^2([0,1])$. Lets write $$ f_n = \sum_{k\in \mathbb{Z}} c_k^{(n)} e^{ikx} $$ where all but finitely many of the $c_k^{(n)}$ are zero (this is because in the span of infinitely many objects we only take a finite number of them to add together)

Now, what I think you are asking is: what can we say about the coefficients $c_k^{(n)}$? The answer is that they converge to the $k$-th Fourier coefficient of $f$ as $n\to\infty$ because $$ \hat f(k) = \langle f, e^{ikx} \rangle = \lim_{n\to\infty} \langle f_n ,e^{ikx}\rangle = \lim_{n\to\infty} c_k^{(n)} $$

In fact if $c_k^{(n)}$ are arbitrary complex numbers, defining $f_n$ as above, we see that $$ \Vert f - f_n \Vert_{L^2} = \sum_{k\in \mathbb{Z}} |\hat f(k) - c_k^{(n)}|^2 $$ assuming convergence. Thus, if $(c_k^{(n)})_k \to (\hat f(k))_k$ as $n\to\infty$ in $\ell^2(\mathbb{Z})$ then $f_n\to f$ in $L^2$, which is a pretty weak condition.

Answer by Helge for Coefficients from Stone Weierstrass versus Fourier Transform

2010-08-24T19:00:23Z

Just a comment if you choose coefficients $c_{k,n}$ such that $$ \lim_{n\to\infty} \left(\sum_{k=-n}^{n} c_{k,n} e^{2\pi i n x}\right) \to f (x) $$ in some sense, e.g. $L^1$, then these are not unique. It is even known that the obvious choice $c_{k,n} = \hat{f}(k) = \int e^{-2\pi i n x} f(x) dx$ is not the best. It's much better to choose $$ c_{k,n} = \left(1 - \frac{|k|}{n}\right) \hat{f}(k). $$ Then one Cesaro sums the Fourier series, and this is known to converge.

As pointed out by Zen Harper below, I should mention that with the choice $c_{k,n} = \hat{f}(k)$ for $-n \leq k \leq n$, the Fourier series of a $L^1$ function must not converge. In fact it can diverge almost-everywhere.

Having said these things, the obvious advantage of this is, that everything is explicit and does not rely on any abstract hocus pocus.

I realized one more thing: Consider the case $f \in L^2$. Then the choice $c_{k,n} = \hat{f}(k)$ for $-n \leq k \leq n$ is optimal. This follows from easy Hilbert space theory!

Answer by Mike Hall for Coefficients from Stone Weierstrass versus Fourier Transform

2010-08-24T22:02:09Z

The Fourier transform is bounded from $L^1([0,1])$ to $\ell^\infty(Z)$ (with norm 1 if you expand in terms of $e^{2\pi ikx}$). If we take a sequence of finite sums of the form $f_n = \sum_k c_{n,k} e^{2\pi ikx}$ where for each $n$ there are only finitely many terms which are not zero ("trigonometric polynomials"), and $\|f_n -f\|_{L^1}\to 0$ , then for all $k$, we have $c_{n,k} \to \hat f(k)$ as $n\to \infty$.

It may help to think of the kind of stuff that doesn't happen in $L^2$: Namely, Kolmogorov famously showed in his first publication that there exist $L^1$ functions $f$ whose Fourier series diverge almost everywhere (and I believe one can arrange for the divergence to be everywhere as well). If such a function $f$ can be written as an infinite sum $f(x) = \sum c_k e^{2\pi ikx}$, with the sum converging in $L^1$ and the coefficients not necessarily the Fourier coefficients, then the sum must converge almost everywhere, so it can't be the Fourier series. By the previous paragraph, it follows that no such representation as a convergent infinite sum is possible.

So what must happen as one uses density of $C(X)$ and Stone-Weierstrass to approximate such a function $f$ by trigonometric polynomials? If $\|f_n - f\|_{L^1} < \epsilon$ , then for all $k$, $|\hat f_n(k) - \hat f(k)| < \epsilon$. One way to think about it is that $\hat f_n(k)$ is forced to be like $\hat f(k)$ whenever the latter is large compared to $\epsilon$, while there's potentially up to an $\epsilon$ of leeway in every coefficient. It is clear that $f_n$ can't just match $f$ in all the Fourier coefficients which are larger than $\epsilon$, though, or else the $L^1$ norm of the resulting sequence would diverge. So density and Stone-Weierstrass are somehow smart enough to use the (less than) $\epsilon$ of room to carry out the $L^1$ approximation.