Proving uniform convergence of $\sum_{n=1}^{\infty}a_n\cos{nx}$ in $\mathbb{R}$

Question

$f(x)$ is a continuous function in $(-\pi, \pi)$, periodic with $2\pi$ period. The derivative, $f'(x)$, is continuous in $[-\pi, \pi]$, except maybe for a finite number of points. $f(x)\sim \frac{1}{2}a_0 + \sum_{n=1}^{\infty}a_n\cos{nx}+b_n\sin{nx}$ is the function Fourier series. I need to prove that $\sum_{n=1}^{\infty}a_n\cos{nx}$ converges uniformly in $\mathbb{R}$ and find its value at each point $x\in\mathbb{R}$

I honestly have no idea how to approach this. I know that if $f$ is continuous then the Fourier series converges uniformly and therefore $\sum_{n=1}^{\infty}a_n\cos{nx}$ also converges uniformly but here $f$ isn't necessarily continuous

EDIT:

I might have found a way prove uniform convergence of $\sum_{n=1}^{\infty}a_n\cos{nx}$

If $\beta_n=\frac{1}{\pi}\int_{-\pi}^{\pi}{f'(x)\sin{nx}\,dx}$ is the Fourier coefficient for $f'(x)$ also \begin{multline} \beta_n = \frac{1}{\pi}\int_{-\pi}^{\pi}{f'(x)\sin{nx}\,dx} = \frac{1}{\pi}\left(f(x)\sin{nx}\bigg|^{\pi}_{-\pi} - \int_{-\pi}^{\pi}{f(x)n\cos{nx}\,dx}\right)\\=-\frac{n}{\pi}\int_{-\pi}^{\pi}{f(x)n\cos{nx}\,dx}=-na_n \end{multline} therefore $$ \sum_{n=1}^{m}{|a_n|} \le \sum_{n=1}^{m}{\frac{1}{n^2}\beta_n^2} \le \sum_{n=1}^{m}{\frac{1}{n^2}\sqrt{\alpha_n^2+\beta_n^2}}\le\sqrt{\sum_{n=1}^{m}{\frac{1}{n^2}}\sum_{n=1}^m{\sqrt{\alpha_n^2+\beta_n^2}}} $$ $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges and $\sum_{n=1}^\infty{\sqrt{\alpha_n^2+\beta_n^2}}$ also converges according to Bessel's inequality for $f'(x)$ hence $\sum_{n=1}^{\infty}{|a_n|}$ converge and from Weierstrass M-test I get that $\sum_{n=1}^{\infty}a_n\cos{nx}$ converge uniformly

EDIT 2:

I found a way to calculate the needed sum:

For $-\pi < x_0 < \pi$ because $f$ is continuous in $x_0$ and the derivative from both sides exists and is finite I get that $f(x_0)=\frac{1}{2}a_0 + \sum_{n=1}^{\infty}a_n\cos{nx_0}+b_n\sin{nx_0}$ hence $$ \frac{f(x_0)+f(-x_0)}{2} = \frac{\frac{1}{2}a_0 + \sum_{n=1}^{\infty}a_n\cos{nx_0}+b_n\sin{nx_0} + \frac{1}{2}a_0 + \sum_{n=1}^{\infty}a_n\cos{(-nx_0)}+b_n\sin{(-nx_0)}}{2}=\frac{a_0+2\sum_{n=1}^{\infty}a_n\cos{nx_0}}{2}\Longrightarrow \sum_{n=1}^{\infty}a_n\cos{nx_0} = \frac{f(x_0)+f(-x_0)}{2}-\frac{1}{2}a_0 = \frac{1}{2}\left(f(x_0)+f(-x_0)-\frac{1}{\pi}\int_{-\pi}^{\pi}{f(x)dx}\right) $$

For $x_1=(2n-1)\pi,\space n\in\mathbb{Z}$ due to the existence of the derivative from both sides I get $$ \frac{1}{2}a_0 + \sum_{n=1}^{\infty}a_n\cos{nx}+b_n\sin{nx} = \frac{f(x_1+0)+f(x_1-0)}{2}\Longrightarrow\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos{nx_1}=\frac{f(-\pi+0)+f(\pi-0)}{2}\Longrightarrow \sum_{n=1}^{\infty}a_n\cos{nx_1} = \frac{1}{2}\left(f(-\pi+0)+f(\pi-0)-\frac{1}{\pi}\int_{-\pi}^{\pi}{f(x)dx}\right) $$ But I'm still unsure about the uniform convergence of the series $\sum_{n=1}^{\infty}a_n\cos{nx}$, if the sum indeed converge uniformly then it converges to a continuous function, and the result I got seems to imply just that.

The way a used in the previous edit seems to apply only when $f$ is continuous but here it might not be.

Answer 1

To adress what is the value of the series, note that since $g(x) = a_0/2 + \sum a_n \cos (nx)$ converges so does $h(x) = \sum b_n \sin (nx)$ and so $f(x) = g(x) + h(x)$ with $g$ even and $h$ odd. Whenever such a decomposition is written, $g$ is the even part of $f$ and $h$ is the odd part of $f$, and can be recovered from $f$ using that $$f(x) + f(-x) = 2g(x)$$ $$f(x)-f(-x) = 2h(x)$$

Answer 2

Look up nice definitions of uniform convergence like Uniformconvergence.

Wikipedia has a page devoted to Convergence of Fourier series. So it is important that $f$ is integrable for development into a convergent Fourier series finite of infinite. Integrable means are continuous on the interval of integration. Steady is not differentiable. To be a differentiable function is a stronger attribute to a function than being continuous. And the derivative being again continuous is stronger than simply being a derivative.

So with the preliminaries of Your problem, there is plenty of space to make a regression to the original convergence problem for the Fourier series.

So there remain the finite number of points. As You write it for the derivative of the function for which the uniform convergence is to be proven. But these finite many points address just the methods with which the integration is accurate. Integration methods like Riemannian or Lebsque usually have this finite point set in spare on which the function might deviate from being continuous or having a unique derivative. These are the so-called zero subspaces. A famous example is the category of stair function. Stair functions are such that there are finite many steps. The integrals of all kinds give the very same result independent of whether the definition is open or closed in the subinterval of the stair function. That is elementary proof in the Analysis. Both Riemann and Lesbeque use stair function prior to exhaustion or axiom of choice.

On the page Mathematical proof there is an overview of the types of proof used here. Deductive reasoning is practiced throughout.

Considering even or odd is not the complete story. There is need to considered a phase.

That far my explanation has an informal character. This changes depending on Your situation. Whether you are in a course or reading a book that has already been introduced as a replacement for the formal background some of the considerations I made so far. Your knowledge background is the empirics from the effects formal repertoire You already have. Having in common the Convergence of Fourier series makes my explanations formal.

One step further is the Carleson's theorem matching the continuous function classes. This pointwise convergence proof follows the uniform convergence proof for $f \in C^{p}$ and $f^(p)$ has a modulus of continuity $\omega$ and is even more rapid. Modulus is a measure of uniform continuity quantitatively. These are formulations for given bounds to fast technically relevant mathematically and numerically calculation. This is much more advance than You need for Your problem.

I hope that my informal proof and the pages I cite convince You that Your effort is not proof. It is the formal production of a formula for calculating the Fourier coefficients and shows that these numerical or symbolic values converge in a practical situation. You do not use the preliminaries of Your problem in them.