Let $f$ be a periodic continuous function on $[0,2\pi]$ with $f'$ continuous. Let the Fourier series of $f$ be ${a_o\over 2} + \sum_{n=0}^\infty(a_n\cos nx + b_n \sin nx).$ Is it true that $$||f-{a_0\over2}||^2 \leq ||f'||$$ , where ||.|| is the $L_2$ norm? I have tried some functions and this seems to hold, but I don't know how to prove it. My guess is to compare the Fourier series of $f$ and $f'$. Let $s_n$ be the n-th partial sum of the Fourier series of $f$, then in $L_2$ $s_n$ converges to $f$. Let $\sum$ be the infinite summation part of the Fourier series ($s_n-{a_0\over2}$) then $$||f-{a_0\over2}|| \leq ||f-s_n||+||\sum||$$ but since $||f-s_n||$ can be made as small as we want, $$||f-{a_0\over2}|| \leq ||\sum||$$.
If I can compare $\sum_{n=0}^\infty(a_n\cos nx + b_n \sin nx)$ and $f'$I can get somewhere....
I think I got it! So I followed user1952009's suggestion. Consider the Fourier series of f': $$ {A_o\over 2} + \sum_{n=0}^\infty(A_n\cos nx + B_n \sin nx)$$ Now, $A_o=\int_0^{2\pi}f'd{x}=f(2\pi)-f(0)=0 $ by periodicity of $f$ $A_n=\int_0^{2\pi}f'\cos{(nx)} dx$ integrating by part, this can be written in terms of f and you find $A_n=-nb_n$. Similarly $B_n=na_n$. $f'$ is Riemann integrable, so its Fourier series converges to $f'$ in $L_2$ norm. Also, $||\sin nx||=||\cos nx||$. Clearly $|a_n|\leq |na_n|$. Therefore $$||f-{a_n \over 2}||^2=||\sum_{n=0}^\infty(a_n\cos nx + b_n \sin nx)||^2=\sum_{n=0}^\infty(|a_n|^2||\cos nx||^2 + |b_n|^2 ||\sin nx||^2) \leq $$ $$\leq\sum_{n=0}^\infty(|na_n|^2||\sin nx||^2 + |nb_n|^2 ||\cos nx||^2= ||f'||^2$$
And the desired inequality follows. Let me know if you think that this works.