The $L^1$ convergence of Fourier series without absolute value

Question

I am having trouble with the following qual problem.

Assume that $f \in L^2[-\pi,\pi]$ and $c_{j}=\frac{1}{2\pi} \int_{-\pi}^{\pi} f(x)e^{ijx}\,dx$.

(a) Prove that $$ \int_{a}^{b} f(x)\,dx = \lim_{n \rightarrow \infty} \sum_{j=-n}^{n}c_{j}\int_{a}^{b}e^{ijx}\,dx $$ for any $[a,b] \subset [\pi,\pi]$.

(b) Does this statement remain true if $f \in L^{1}[-\pi,\pi]$?

I solved the part $(a)$. The following is the proof: to show the equality, it suffices to show the following equality. $$ \lim_{n \rightarrow \infty} \left|\int_{a}^{b} (f(x) - \sum_{j=-n}^{n}c_{j}e^{ijx})\,dx\right| = 0. $$ However, by the $L^2$ convergence of Fourier series $$ \begin{split} \left|\int_{a}^{b} (f(x) - \sum_{j=-n}^{n}c_{j}e^{ijx})\,dx\right| &\leq \int_{a}^{b} |f(x) - \sum_{j=-n}^{n}c_{j}e^{ijx}|\,dx \\& \leq c\int_{-\pi}^{\pi} |f(x) - \sum_{j=-n}^{n}c_{j}e^{ijx}|^2\,dx \rightarrow 0. \end{split} $$ This completes the proof of part $(a)$.

Part $(b)$ seems to be more difficult. I know several theorems about Fourier series of $f \in L^1$. The first theorem is that there exists a function $f \in L^1$ whose Fourier series diverges almost everywhere. Another theorem is that the partial sum of Fourier series of $f$ does not necessarily converge to $f$ in $L^1$-norm. None of them seems to directly apply to our problem. Moreover, if $[a,b]=[-\pi,\pi]$, the equality holds true for any $f \in L^1$.

Do you have any ideas about part $(b)$? Any comments are welcomed.

Answer 1

Hints: let $g(x)=\int_0^{x} f(t)\, dt$. Then $g$ is a continuous functions of bounded variation. Hence its Fourier series converges uniformly. Now compute the Fourier coefficients of $g$ using integration by parts and conclude the proof. A more general result: let $\mu$ be a complex measure such that $\mu \{x\}=0$ for all $x$ and consider the Fourier series $\sum e^{inx} \hat {\mu} (n)$ where $\hat {\mu} (n) =\int e^{-inx} \, d \mu(x)$. Then the 'integrated series' $\sum (\int_0^{x} e^{int} \, dt) \hat {\mu} (n)$ converges uniformly. The proof is almost the same as the one in the earlier case.

Answer 2

Yaoliding gave me the proof of the part $(b)$. Suprisingly, it turned out to be true. The idea is to use the kernel expression: $$ \sum_{j=-n}^{n}c_{j}e^{ijx} = (f * D_{n})(x), $$ where $$ D_{n}(x)=\frac{\sin((n+1/2)x)}{\sin(x/2)}. $$ With this and some property of convolusion, $$ \int_{a}^{b} \sum_{j=-n}^{n}c_{j}e^{ijx}\, dx = \int_{-\pi}^{\pi} \chi_{[a,b]}(x)(f*D_{n})(x)\, dx = \int_{-\pi}^{\pi} (\chi_{[a,b]}*D_{n})(x)f(x)\, dx. $$ We need some basic properties of Dirichlet kernel.

1. $(\chi_{[a,b]}*D_{n})(x)$ is uniformly bounded on the interval $[-\pi,\pi]$

2. $(\chi_{[a,b]}*D_{n})(x)$ converges to $\chi_{[a,b]}(x)$ at every continuous point except.

We now go back to our original problem. \begin{gather} \left|\int_{a}^{b}(f(x) -\sum_{j=-n}^{n}c_{j}e^{ijx})dx \right| = \left| \int_{-\pi}^{\pi}f(x)(\chi_{[a,b]}(x)-(D_{n}*\chi_{[a,b]})(x))\,dx \right|. \end{gather} We divide the interval $[-\pi,\pi]$ into two pieces: $$E_{1}=[-\pi,a-\epsilon]\cup [a+\epsilon,b-\epsilon] \cup [b+\epsilon, \pi], \\ E_{2}=[a-\epsilon,a+\epsilon] \cup[b-\epsilon,b+\epsilon].$$ Note that the set $E_{1}$ is compact. The integral over the set $E_{1}$ is small enough because of the property $(2)$, and the integral over the set $E_{2}$ is also small enough because of the property $(1)$. Therefore, we get the part $(b)$.