Baby Rudin's Theorem 8.14 and convergence of Fourier series

Question

I need to show that the fourier series of the function $$f(x)=(\pi-|x|))^2$$ converges to the function $\forall x \in[-\pi,\pi] $.

Now, Baby Rudin's Theorem 8.14 says if for some $x$ there exists a $\delta > 0 $ and $M<\infty$ such that $|f(x+t)-f(x)|\leq M|t|\hspace{0.5cm} \forall t \in (-\delta,\delta)$, then the fourier series will converge to the function at that $x.$ ($f$ is Riemann integrable)

My questions are

1. I am viewing the above property as Locally Lipschitz condition. Is that correct?
2. To show that the function $f(x)=(\pi-|x|))^2$ abides with the above property, I am trying to bound $|f^\prime| \forall x \in[-\pi,\pi] $ but $|f^\prime|$ does not exist at zero.How do I solve this question then? Any hints are welcome. Thanks in advance.

Answer 1

1. YES
2. No need to use derivatives or MVT. Use just simple High School algebra. Take $\delta =1$ and $M=1+2|x|+\pi$. We have $|f(x+t)-f(x)|=|(\pi^{2}+(x+t)^{2}-\pi |x+t|)-(\pi^{2}+x^{2}-\pi |x|)|$. Using the fact that $|x|-|x+t| \leq |t|$ show that $|f(x+t)-f(x)| \leq M|t|$.