Fourier Series Convergence

Question

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function that is differentiable at the point $x_0$. Prove that $S_n(f(x_0))$ converges to $f(x_0)$, where $S_n$ denotes the partial sums of the Fourier series.

Note: We only covered the Riemann Lebesgue Lemma for the hypotheses that $f$ is continuous.

Attempts: We know that $S_n(f(x_0))-f(x_0)=\frac{1}{2\pi}\int_{-\pi}^{\pi}[f(x_0-y)-f(x_0)][D_N(y)]dy \\=\frac{1}{2\pi}\int_{0}^{\pi}[f(x_0-y)+f(x_0+y)-2f(x_0)][D_N(y)]dy$

where $D_n(y)$ is the Dirichlet kernel. I'm not sure how to continue from here. I've tried dividing the interval $[-\pi,\pi]$ into three different subintervals: $[-\pi,-\delta]$, $[-\delta,\delta]$, $[\delta, \pi]$ and looking at how the function behaved in each subinterval, but I couldn't find a way to bound $S_n(f(x_0))-f(x_0)$ in terms of $n$. Any help would be much appreciated.

Answer 1

Note $\lim_{y\to 0}\,(f(x_0-y)-f(x_0))/\sin(y/2)$ exists since $f'(x_0)$ exists. So we can regard this function as continuous on $[-\pi,\pi].$ We're integrating this against the numerator in $D_n(y),$ which is $\sin (n+1/2)y.\,$ Apply Riemann-Lebesgue.

Answer 2

The difference between the partial sum and $f(x)$ (I drop the subscript here) is given as you have given by

${t_n}(x) - f(x) = = \frac{1}{\pi }\int_0^\pi {\left( {f(x + u) + f(x - u) - 2f(x)} \right){D_n}(u)du} $

$ = \frac{1}{\pi }\int_0^\pi {\left( {f(x + u) + f(x - u) - 2f(x)} \right)\frac{{\sin \left( {(n + {\textstyle{1 \over 2}})u} \right)}}{{2\sin ({\textstyle{1 \over 2}}u)}}du} $

$ = \frac{1}{\pi }\int_0^\pi {\frac{{\left( {f(x + u) + f(x - u) - 2f(x)} \right)}}{u}\frac{u}{{2\sin ({\textstyle{1 \over 2}}u)}}\sin \left( {(n + {\textstyle{1 \over 2}})u} \right)du} $

Now

$\mathop {\lim }\limits_{u \to 0} \frac{{\left( {f(x + u) + f(x - u) - 2f(x)} \right)}}{u}\frac{u}{{2\sin ({\textstyle{1 \over 2}}u)}}$

$ = \left( {\mathop {\lim }\limits_{u \to 0} \frac{{\left( {f(x + u) - f(x)} \right)}}{u} + \mathop {\lim }\limits_{u \to 0} \frac{{\left( {f(x - u) - f(x)} \right)}}{u}} \right)\mathop {\lim }\limits_{u \to 0} \frac{u}{{2\sin ({\textstyle{1 \over 2}}u)}}$

$ = \left( {f'(x) - f'(x)} \right) \cdot 1 = 0$ .

Therefore,

$\frac{{\left( {f(x + u) + f(x - u) - 2f(x)} \right)}}{u}\frac{u}{{2\sin ({\textstyle{1 \over 2}}u)}}$ is Lebesgue integrable. Hence by the Riemann-Lebesgue Lemma,

$\frac{1}{\pi }\int_0^\pi {\frac{{\left( {f(x + u) + f(x - u) - 2f(x)} \right)}}{u}\frac{u}{{2\sin ({\textstyle{1 \over 2}}u)}}\sin \left( {(n + {\textstyle{1 \over 2}})u} \right)du} $

tends to $0$ . That means ${t_n}(x) - f(x)$ tends to 0.