Let $ f:\mathbb{R} \to \mathbb{C} $ be periodic function and continuous in the interval $ [-\pi,\pi) $. (The period is $2\pi $ ).
We are given that $ lim_{x \to \pi^-} f(x) \neq f(-\pi) $, prove that $ \sum_{k \in \mathbb{Z}} | \hat{f}(k) | = \infty $
.
I tried to assume for contradiction that the sum actually converge, then by the M-test we'll get that Fourier series of $ f $ converge, and thus converge to $f$ in any point $ x_0 $ where $ f $ is continuois at. But Im not sure how to get the contradiction.
Ive been thinking about it for a while, any ideas would help.
Thanks in advance
Best Answer
If $\sum |\hat {f} (k)| <\infty$ then $g(x)=\sum \hat {f} (k) e^{ikx}$ defines a continuous function $g$ with $g(x+2\pi)=g(x)$ for all $x$. Note that the series is uniformly convergent. You an calculate $\int g(x)e^{-ikx} dx$ by integrating term by term. You will find that $f$ and $g$ have the same Fourier coefficients. This implies that $f=g$ almost everywhere. By continuity we get $f(x)=g(x)$ for all $x <\pi$. But $g(\pi)=g(-\pi)$ so $\lim_{x \to \pi -} f(x)=\lim_{x \to \pi -} g(x)=g(\pi)=g(-\pi)=f(-\pi)$.