Let $f(x)$, a $2\pi$ periodic funciton such that $f(0) = 1$ and for every $0\ne x\in[-\pi,\pi]$: $f(x) = 1 + \sin \frac{\pi^2}{x}$. Is the Fourier series of $f(x)$ converges at $x=0$? If so, what it's value there?
Well, $\lim_{x\to 0} f(x)$ doesn't exist. We have learned in class that if a function is continuous at a point $x_0$ (or even locally Lipchitz) then it's Fourier series converges to $f(x)$. Moreover, we have learned that if the sided limits exist but different then the Fourier series converges to $$\frac{f(x^+)+f(x^-)}{2}.$$
But here, the limit doesn't even exist. So is it implying the Fourier series doesn't converge to $f(0)$?
Best Answer
No, it doesn't imply that, and in this case, the Fourier series of $f$ does converge to $f$ pointwise.
If we look at $g(x) = f(x) - 1$, we note that $g$ is an odd function, hence in the real Fourier series, only sine terms occur, and of course $\sin (nx) = 0$ for $x = 0$, whence the Fourier series of $g$ converges to $g$ at $0$. For $0 < \lvert x\rvert < \pi$, the continuous differentiability of $g$ in a neighbourhood of $x$ implies the (pointwise) convergence of the Fourier series, and at $\pm \pi$, we have $g(\pm\pi) = 0$, and $\sin (\pm n\pi) = 0$ for all $n$, so pointwise convergence there too (that also follows from $g$ being locally Lipschitz in a neighbourhood of $\pm\pi$).
Adding the constant term $\cos (0x)$ to get the Fourier series of $f$ doesn't change the convergence behaviour.