Let $f(x)$ be the $2π$-periodic function defined by $ f(x)= \begin{cases} 1+x&\,x \in \mathbb [0,π)\\ -x-2&\, x \in \mathbb [-π,0)\\ \end{cases} $

Question

Let $f(x)$ be the $2π$-periodic function defined by

$ f(x)= \begin{cases} 1+x&\,x \in \mathbb [0,π)\\ -x-2&\, x \in \mathbb [-π,0)\\ \end{cases} $ Then the Fourier series of $f$

$(A)$ converges to $-1/2$ in $x=0$ $(B)$ does not converge pointwise in all points where $f$ is discontinuous $(C)$ does not converge in $x=π/2$ $(D)$ converges to $1$ in $x=0$

Picture of actual question is Link to the Real Question

Correct answer is 'A' but I am unable to get it. I have tried calculating the Fourier series and then calculate for its convergence. The calculations are difficult to be shown here as I am not acquainted with the Math Latex. Please help me solve the question

Answer 1

A basic fact about the Fourier series of a $2\pi$-periodic piecewise $C^1$ function $f$ is that it converges pointwise to $$f_*(x):=\frac{f(x^-) + f(x^+)}{2},$$ where $f(x^\pm) = \lim_{t \to x^\pm} f(t)$ are the left- and right-hand limit of $f$ at $x$; this holds for all $x$ in the domain of $f$, even where $f$ is not continuous. Where $f$ is continuous, $f_* = f$ (and the convergence is uniform).

Answer 2

The word "at" is more appropriate than "in" here.

For a piecewise continuous function, at points of discontinuity, the Fourier series converges to the midpoint of the limits of each piece.

The first piece has $\lim_ {x\downarrow 0} f(x) = f(0)=1$ and the second piece has $\lim_ {x\uparrow 0} f(x) =-2.$

The midpoint is $\frac{(1-2)}{2} = - \frac{1}{2}.$

So yes, choice (A) is correct.