The limit of the convergent Fourier series of $f$ is $f$

Question

I have to try and solve the following question:

Let $f$ be a continuous function on $[-\pi,\pi].$ Suppose that the Fourier series converges uniformly. Then its limit must be $f.$

MY ATTEMPT

Say $\sum_{n=-N}^N \hat{f}(n)e^{in\theta}$ converges uniformly to $L(\theta).$ Then,
$$\sum_{n=-N}^N r^{|n|}\hat{f}(n)e^{in\theta}$$ converges uniformly for $r\rightarrow 1^-$ to $L(\theta).$

Since $f$ is continuous, by Poisson's Theorem we have that $\sum_{n=-N}^N r^{|n|}\hat{f}(n)e^{in\theta}$ converges uniformly to $f(\theta)$ as $r\rightarrow 1^-.$ So $L(\theta)=f(\theta),$ for all $\theta.$

---

I'm worried about a logical flaw in my attempt or an overlook of some property of uniform convergence. Also, the exercise comes with the hint:

HINT: Calculate the Fourier coefficients of the limit function.

Now, I haven't used this idea my attempt and really would like some help getting this approved/rectified, I would really appreciate any help! Thank you so much!

P.S. Going with the hint, I thought maybe I could use the fact that if all the Fourier coefficients are the same for two functions, then the functions are identical at all points; I ended up at a dead-end here though.

Answer 1

If the Fourier Series for $f$ converges uniformly to $g$, then $g$ is continuous and \begin{align} \hat{g}(n)&=\frac{1}{2\pi}\int_{0}^{2\pi}g(x)e^{-inx}dx \\ &=\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}\sum_{k=-K}^{K}\hat{f}(k)e^{ikx}e^{-inx}dx \\ &= \lim_{N\rightarrow\infty}\sum_{k=-K}^{K}\hat{f}(k)\frac{1}{2\pi}\int_0^{2\pi}e^{i(k-n)x}dx=\hat{f}(n). \end{align} Therefore, the Parseval identity gives $\int_0^{2\pi}|f(x)-g(x)|^2dx = 0$. So, the following holds for all $y$ $$ 0 \le \int_0^y|f(x)-g(x)|^2dx \le \int_0^{2\pi}|f(x)-g(x)|^2dx =0. $$ It follows that $\int_0^y|f(x)-g(x)|^2dx =0$ for all $0\le y\le 2\pi$. By the Fundamental Theorem of Calculus, the derivative with respect to $y$ is $|f(y)-g(y)|^2$, which must be $0$. Therefore $f=g$.