Let's say I have a sequence of piece-wise continuous, square-integrable functions $$u_{n}(x)\in L_{pc}^{2}(- \infty ,\infty )$$ I am assuming the series $S(x)=\sum_{n=1}^{\infty}u_{n}(x)$ converges in $L_{pc}^{2}(- \infty ,\infty )$.
Is it true that $$F(S(x))=\sum_{n=1}^{\infty}F(u_{n}(x)) $$ meaning, can I under the above conditions, interchange the summation with the fourier operator?
My thought is yes. I have tried proving it in the following way:
Since $\sum_{n=1}^{\infty}u_{n}(x)$ converges in $L_{pc}^{2}(- \infty ,\infty )$ by definition it means the partial sums sequence $S_{N}(x)=\sum_{n=1}^{N}u_{n}(x)$ converges and thus a cauchy sequence. By Plancherel's theorem we can conclude that $\hat{S}_{N}(x)=\sum_{n=1}^{N}\hat{u}_{n}(x)$ is a cauchy sequence as well.
Because $L_{2}(-\infty,\infty)$ is complete, the cauchy sequence $\hat{S}_{N}(x)$ converges to some element in the space, we'll denote it as $\hat{S}(x)$.
What I am missing here is how can I conclude that $\hat{S}(x)$ is really the fourier transform of $S(x)$ ?
If $\{ h_n \}$ converges in $L^2$ to $h$, then $\mathcal{F}h_n$ converges in $L^2$ to $\mathcal{F}h$ because $$ \|\mathcal{F}h_n-\mathcal{F}h\|=\|\mathcal{F}(h_n-h)\|=\|h_n-h\|. $$ Take $h_n = \sum_{k=1}^{n}u_k$ and $h=\sum_{k=1}^{\infty}u_k$. Because $\{ h_n \}$ converges in $L^2$ to $h$, then $\mathcal{F}h_n = \sum_{k=1}^{n}\mathcal{F}u_k$ converges in $L^2$ to $\mathcal{F}h$. I think that's what you're asking. The sum of Fourier transforms is interpreted in an $L^2$ sense.