fourier series expansions

Question

If someone could help me out with this problem and/or direct me to a proof somewhere I would appreciate it. Is there a name for such a proof so that I can look it up? I tried bounds on norm of Fourier Series and other similar searches.

Answer 1

Rough Sketch: Naively, you have \begin{align} f- f_N = \frac{1}{\sqrt{2\pi}}\sum_{|n|\geq N} \hat f(n)e^{inx}. \end{align} In particular, the $j$th derivative is given by \begin{align} \frac{d^j}{dx^j}(f-f_N) = \frac{1}{\sqrt{2\pi}}\sum_{|n|\geq N} (in)^j\hat f(n)e^{inx} \end{align} which means \begin{align} \left\|f-f_N \right\|^2_{L^2([0, 2\pi])}+\left\| \frac{d^j}{dx^j}(f-f_N)\right\|^2_{L^2([0, 2\pi])} =& \sum_{|n|\geq N} (1+n^{2j})|\hat f(n)|^2\\ =&\ \sum_{|n|\geq N} (1+n^{2j})(1+n^{2(s-j)})(1+n^{2(s-j)})^{-1}|\hat f(n)|^2\\ \leq&\ C\sum_{|n|\geq N} (1+|n|^{s-j})^{-2}(1+n^{2s})|\hat f(n)|^2 \\ \leq&\ C(1+N^{s-j})^{-2}\sum_{|n|\geq N} (1+n^{2s})|\hat f(n)|^2 \end{align} since $|n| \geq N$.

Additional Remark: Let us note for $x, y\geq 1$ \begin{align} (1+x)(1+y) = 1+x+y+xy \leq 3(1+xy). \end{align}