Let $f:\mathbb{T} \to \mathbb{C}$ be a 1-periodic continuous function whose Fourier coefficients satisfy $$\hat{f}(j) = 0$$ for all $j$ not of the form $$\pm2^k \text{ for some } k \in \mathbb{N} \cup \{0\}.$$ Show that the Fourier series of $f$ converges to $f$ for all $x$.
I know two continuous functions are the same if they have the same Fourier coefficients but I know nothing about $f$ except when its Fourier coefficients are zero.
I tried to see that for all $\epsilon>0$ there is an $N$ such that $n\geq N$ implies $$|f(x)-\sum_\limits{k=2^j}^n \hat{f}(k)e^{2k\pi i x}|<\epsilon$$ but I can't see how to do this.
I wonder if it is important that we have an exponential, would the same holds if $\hat{f}(j)=0$ for all odd $j$ for example?
Nice problem, very specific to the sequence $(2^k)$.
$f$ is continuous $1$-periodic and in $L^2([0,1]),L^1([0,1])$ we have $$f = c_0+\sum_{k=0}^\infty (c_k e^{2i \pi 2^k x}+c_{-k} e^{-2i \pi 2^k x})$$
Then look at $$f_n(x) =f(x)+c_0- \frac{1}{2^n} \sum_{m=0}^{2^n-1} f(x+\frac{m}{2^n})$$
What is its Fourier series ?