Define the Fourier coefficients $\hat{f}(n)$ of a function $f\in L^2(T)$, $T$ is a unit circle, by:
$\hat{f}(n)=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(e^{i\theta})e^{-in\theta}d\theta$
Let $\Lambda_nf=\sum\limits_{k=-n}^{n}\hat{f}(k)$. Prove that the set $A=\{f\in L^2(T):\lim\limits_{n\to\infty}\Lambda_nf\, \text{exists} \}$ is a dense subspace of $L^2(T)$ of the first category.
My thoughts: If we take $f(\theta)=e^{ik\theta}$, $k\in\mathbb{Z}$, then the before mentioned limit exists. But $\{e^{ik\theta}:k\in\mathbb{Z}\}$ is dense in $L^2(T)$, so $A$ is dense too. Is this true? Any hints about the second part (of the first category) are welcome. Thank you very much!
Edit: there is a theorem which states: Suppose $X$ and $Y$ are topological vector spaces and $\Lambda_n$ is a sequence of continuous linear mappings from $X$ into $Y$. If A (set where limit exists) is of the second category in $X$ and if $Y$ is an $F$-space, then $A=X$ and $\Lambda f=\lim\limits_{n\to\infty}\Lambda_n f$ is continuous.
If I find $g\in L^2$ for which the limit doesn't exist, that would be it, right? Because then we could suppose that $A$ is not of the first category (i.e. it is of the second category) and we could apply this theorem. The conclusion would be that $A=X$, but that would contradict the found $g$.
Your statement about the set $\exp(i k \theta)$ being dense is false. However, the set of their linear combinations (a.k.a trigonometric polynomials) IS dense (since continuous functions are dense), so your set $A$ is dense also. The rest of the argument seems correct, though I am not sure what an $F$-space is.