Using Weistrass Approximation Theorem to define fourier series convergence.

Question

Weistrass Approximation Theorem: Let f be continuous on [-$\pi$,$\pi$] with $f(-\pi)=f(\pi)$. Then for each $\epsilon>0$ there is a trigonometric polynomial T such that $|f(x)-T(x)|<\epsilon$ for all x an element of $[-\pi,\pi]$.

We were also given a theorem saying that if $f(x)$ is square integrable then $\int[f(x)-T(x)]^2dx$ is miniimised when $T(x)$ is equal to the functions fourier series.

Now, we know that this implies convergence of the functions fourier series to the function itself in the $L^2$ sence (Lebesgue norm), but what I dont understand is why this doesn't imply uniform convergence of the functions fourier series to the function itself. I am aware that you do need stronger conditions to make this claim, but I dont see how Weistrass' theorem is not showing uniform convergence, and hence why we can't just use this theorem to say that uniform convergence holds.

Any help would be greatly appreciated, thanks!

Answer 1

The Weierstrass theorem you quote requires $ f $ continuous and with equal values at the endpoints Neither condition is required to be square integrable.

Answer 2

I don't completely understand your question, let me just state some facts which might be related.

First, Weistrass Approximation Theorem states that if $f$ is continuous on $[-\pi, \pi]$ and $f(-\pi) = f(\pi)$, then for all $\epsilon>0$, there is $T$ so that $|f(x) -T(x) |<\epsilon$. Thus

• $T$ depends on $\epsilon$,
• $T$ is not unique given a fixed $\epsilon$.
• $T$ in general is not the fourier series $S_n(f)$ for any $n$.