$\{f\in L^2(S^1):\lim_{n\rightarrow\infty}\Lambda_n(f)\text{ exists}\}\subseteq L^2(S^1)$ is a dense subspace of first category.

Question

Let $f\in L^2(S^1)$ and $\hat{f}(k)$ be the Fourier coefficient of $f$ at $k\in\mathbb{Z}$. Define
\begin{equation*}
\Lambda_n(f):=\sum_{k=-n}^{n}\hat{f}(k).
\end{equation*}

Then $\{f\in L^2(S^1):\lim_{n\rightarrow\infty}\Lambda_n(f)\text{ exists}\}\subseteq L^2(S^1)$ is a dense subspace of first category.

Edited: We let $f_m(\theta)=e^{im\theta}.$ Then the Fourier coefficient is
\begin{equation*}
\begin{split}
\hat{f}_m(k)&=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)e^{-ik\theta}d\theta\\
&=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{im\theta}e^{-ik\theta}d\theta\\
&=\begin{cases}0&\text{when } m\neq k\\1&\text{when } m=k\end{cases}
\end{split}
\end{equation*}
Thus for each $m\in\mathbb{Z}$ the function $f_m(\theta)$ has the property that $\lim_{n\rightarrow\infty}\Lambda_n(f_m)$ exists.

Now we let $f:=\sum_{j=-N}^{N}c_jf_j$. Then
\begin{equation*}
\hat{f}(k)=\sum_{j=-N}^{N}c_j\hat{f}_j(k)=c_k\hat{f}_k(k)=c_k.
\end{equation*}
By definition
\begin{equation*}
\Lambda_n(f)=\sum_{k=-n}^{n}\hat{f}(k)=\sum_{k=-n}^n c_k.
\end{equation*}
It follows that
\begin{equation*}
\lim_{n\rightarrow\infty}\Lambda_n(f)=\lim_{n\rightarrow}\sum_{k=-n}^nc_k.
\end{equation*}

My question is why $\lim_{n\rightarrow\infty}\Lambda_n(f)$ exists? In comment section Nate Eldredge suggested to show that the collection E of all such
f is an algebra of continuous functions that separates points. Then conclude that E is dense in $C(S_1)$ and therefore, by standard arguments, also dense in $L^2(S^1)$. But I don't know how. Any help is greatly appreciated.

Answer 1

That this set, let's call it $E,$ is a subspace of $L^2$ is clear.

Density: Each $e^{ikt}$ is in $E$ since $\Lambda_n (e^{ikt})$ is constant for $|n|>k.$ As $E$ is a linear subspace, it therefore contains all trigonometric polynomials. These polynomials are dense in $L^2,$ hence so is $E.$

For $N\in \mathbb N,$ let $F_N=\{f\in L^2: |\Lambda_n f| \le N, n=0,1,\dots\}.$ Then $E$ is contained in $\cup_{N=0}^\infty F_N.$ So it is enough to show each $F_N$ is closed and nowhere dense.

Suppose $f_1,f_2,\dots \in F_N$ and $f_m \to f$ in $L^2.$ Then for each fixed $n,$

$$|\Lambda_n f| = \lim_{m\to \infty} |\Lambda_n f_m| \le N.$$

Thus $f \in F_N,$ proving $F_N$ is closed.

Now let $g= \sum_{k=1}^{\infty}\dfrac{e^{ikt}}{k}.$ Then $g\in L^2.$ Let $f\in F_N.$ Then as $\epsilon\to 0^+,$ $f+\epsilon g\to f$ in $L^2.$ But

$$|\Lambda_n(f+\epsilon g)|\to \infty$$

as $n\to \infty.$ This shows $F_N$ contains no open ball $B(f,r)$ in $L^2$ of positive radius. I.e., $F_N$ is nowhere dense.