Prove or disprove the following:
Let $V$ be the space of continuous functions on $[0,2\pi]$ with $f(0)=f(2\pi)$. Let $\|f\|=\text{sup}\{|f(x)|: 0\leq x \leq 2\pi\}$. Let $N$ be a positive integer and $M$ be the subspace of $V$ spanned by the functions $\{e^{-inx}: -N\leq n \leq N\}$. Show that $\text{inf}\{\|f-g\|: g \in M\}=\|f-h\|$ where $$h(x)= \sum_{n=-N}^{N} f\hat(n)e^{inx}$$
And $f\hat(n)$ is the usual Fourier coefficient.
Take $N=0$ and the piecewise linear function with $f(0)=f(2\pi)=1$ and $f(\frac{\pi}{2})=f(\frac{3\pi}{2})=-1$. Its graph looks like \__/.
The best uniform approximation by an element of $M$ is the $0$ function, but $$ \hat{f}(0) = -\frac{1}{2} $$
Actually the same idea works for any $N \in \Bbb N$. The function $$ f(x) = \begin{cases}1 &\text{if} & x \in2\pi\Bbb Z\\-1& \text{else}\end{cases} $$ is not continuous but can be approximated by functions $f_\epsilon$ of the type \__/ with $\|f-f_\epsilon\|_{L^1}\leq \epsilon$. In particular $|\widehat{f_\epsilon}(n)-\widehat{f}(n)| \leq \epsilon$ for any $n$ with $$ \widehat{f}(n) = \begin{cases}-1 & \text{if}& n=0\\0 & \text{else}\end{cases} $$
For $\epsilon$ small enough (for instance $\epsilon < 1/(2N+1)$), we have $$ \|f_\epsilon - S_Nf\|_\infty \geq \|f_\epsilon - (-1)\|_\infty - 2(N+1)\epsilon) = 2 - (2N+1)\epsilon > 1= \|f_\epsilon - 0\|_\infty $$
Of course, the $0$ function need not be the best uniform approximation of $f_\epsilon$ in $M$, but it is already better than the Fourier trigonometric polynomial.
This is not true! While it is true that, there is a unique best approximant (in $M$, the space of trigonometric polynomials of degree atmost $n$) to any continuous $2\pi$-periodic function, it is not true that this function is $h \equiv S_n(f)$. In fact, if $E_n(f)$ denotes the error in approximating $f$ by the unique best approximant, then,
Theorem: With the notations as above,
$$\|f-S_n(f)\|_\infty \leqslant (3+\log n)E_n(f)$$
Edit:
The proof is very simple: one needs to know a decent bound for the integral of the Dirichlet Kernel over a period.
I think I read this from here.
