let $T\subseteq [-\pi,\pi]$ be any countable set . show that there is a $f\in C(\mathbb{T})$ such that $S_N f(x)$ diverge at each $x\in \mathbb{T}$
i know that that there exist $g\in C(\mathbb{T})$ such that $S_Ng(0)$ diverge.
where $S_Nf(x)$ denote sequence of partial sums of fourier series of $f$.
how to approach this problem.do we need to use the result stated above or we have find such $f$ by some other mean.
any hint please. thanks in advanced .
I don't see how divergence at one point is going to get you divergence at every point of a countable set. But the standard example of a divergent Fourier series actually gives this:
This implies immediately that
And that fact does immediately extend to unboundedness at every point of an arbitrary countable set.
The idea is very simple. Say $M$ is the maximal function $$Mf(x)=\sup_N|S_N(f)(x)|.$$Let $(n_j)$ be a sequence that contains every positive integer infinitely many times, for example $(1,1,2,1,2,3,1,2,3,4,1\dots)$. Say $T=\{x_n\}$.
Choose trigonometric polynomials $P_j$ with $||P_j||<1/2^j$ and $$MP_j(x_{n_j})|>j+\sum_{k=1}^{j-1}||MP_j||_\infty.$$
Choose $(M_j)$ so that the coefficients of $e^{iM_jt}P_j(t)$ are supported in disjoint intervals, and let $$f(t)=\sum_je^{iM_jt}P_j(t).$$It follows that $$Mf(x_n)=\infty$$for every $x_n\in T$. I'll let you figure out why - I did the hard part...
Edit: I just realized that it may not be clear how $MP_j$ large can imply $Mf$ large, since the symettric partial sums for $P_j$ don't come up in the partial sums for $f$. But what I said above works fine, using the following corollary to the second lemma above: