I'm getting a little confused about the next exercise:
Suppose that $f_{k}\in L^1[-\pi,\pi]$ and $f\in L^1[-\pi,\pi]$ are such that $f_k \to f$ when $k\to \infty $.
Prove that $\hat{f_k} \to \hat{f}$ uniformly, where $\hat{f_k}$ and $\hat{f}$ are the Fourier coefficents of $f_k$ and $f$ respectively.
Mi attempt was:
For an $\varepsilon>0$ we know that $\exists N \in \mathbb{N}$ such that $|f_k(\theta)-f(\theta)|<\varepsilon$ for every $k>N$ and $\theta\in[-\pi,\pi]$. Then, notice the following:
$\hat{f_k}-\hat{f}=\frac{1}{2\pi}\int_{-\pi}^{\pi} f_k(\theta) e^{-in\theta}dx-\frac{1}{2\pi}\int_{-\pi}^{\pi} f(\theta) e^{-in\theta}dx=\frac{1}{2\pi}\int_{-\pi}^{\pi} [f_k(\theta)-f(\theta)] e^{-in\theta}dx\frac{1}{2\pi}\leq\frac{1}{2\pi}|\int_{-\pi}^{\pi} [f_k(\theta)-f(\theta)] e^{-in\theta}dx|\leq\frac{1}{2\pi}\int_{-\pi}^{\pi} |f_k(\theta)-f(\theta)| e^{-in\theta}dx\leq\frac{1}{2\pi}\int_{-\pi}^{\pi} \varepsilon e^{-in\theta}dx=\frac{\varepsilon}{2\pi}\int_{-\pi}^{\pi} e^{-in\theta}dx$
My problem is that the final integral is equal to $\frac{2sin(n\pi)}{n}=0$ since $n\in\mathbb{N}$, and I don't like that. Do you see any mistake? or Can you see another way to prove this?
Any help is appreciated, thank you.
Your inequalities don't make sense. The argument is quite simple: $|\int f_n(t)e^{-int} dt-\int f(t)e^{-int}dt|\leq \int |f_k(t)-f (t)||e^{-int}|dt \to 0$ since $|e^{-int}|=1$ for all $t$ and $n$.
[You did not in what sense $f_k \to f$. I am assuming that $\int |f_n-f| \to 0$].