Convergence of Fourier series in Sobolev space

Question

So the problem is if $f\in H^{\frac{1}{2}}([0,1])\cap C([0,1])$, then $S_Nf$, the partial sum of fourier series converges uniformly to $f$. How would you show this by considering the quantity $|S_Nf-\sigma_Nf|$. We know that $\sigma_Nf\rightarrow f$ uniformly. More over, $$\sum_{n\in \mathbb{Z}}|n||\widehat{f}(n)|^2$$ thanks

Answer 1

You have the right approach. Given $$\sum_{n=-\infty}^\infty |n\hat f(n)^2|<\infty$$ we need to show the $\ell^1$ difference tends to zero: $$\frac 1 N\sum_{n=-N}^N |n\hat f(n)|\to 0\qquad\text{ $(N\to\infty)$.}$$

The rest is just epsilons, deltas, and Cauchy-Schwarz.

Given $\epsilon>0,$ pick $N_0$ such that $\sum_{|n|\geq N_0} |n\hat f(n)^2|<\epsilon^2/8.$ Since $N_0$ is fixed, for sufficiently large $N$ we have $$\frac 1 N\sum_{|n|\leq N_0} |n\hat f(n)|\leq \epsilon/2.$$ And using Cauchy-Schwarz we get $$\frac 1 N\sum_{N_0\leq |n|\leq N} |n\hat f(n)| \leq \frac 1 {N^{1/2}}\sum_{N_0\leq |n|\leq N} |n\hat f(n)^2|^{1/2} \stackrel{\text{C-S}}{\leq} \frac 1 {N^{1/2}} \sqrt{(\epsilon^2/8) 2N}\leq \epsilon/2$$

Putting this together gives the result.