$f_m:\mathbb{R} \rightarrow \mathbb{R}$ is function series that are integrable on any compact segment , $2\pi$ periodic and $f_m \overset{u}{\rightarrow} f$ converging uniformly. Moreover for all $n \in \mathbb{Z} \setminus \{0\}$ and all $m\ge 1$ $$|\hat{f_m}|\le \frac{1}{|n|^3}$$
- Is f necessarily continuously differentiable under given conditions? prove or give a counter example.
- now you are given that $f_m$ is continuous for all $m$ is f necessarily continuously differentiable now ? prove or give a counter example.
since we are not given that $f_m$ is continuous in first question i'm trying to use it to "build" a counter example for the first question .
but i can't understand why are we given that information about fourier coefficients and how to use it .
have no clue on the second question and in general how one can prove a function is continuously differentiable if the function itself is not known.
I will assume that the inequality is $$ |\widehat{(f_m)}_n|\le\frac{1}{n^3}\quad\forall n\in\mathbb{Z}. $$ Since $f_m\to f$ uniformly, it follows that $$ |\hat f_n|\le\frac{1}{n^3}\quad\forall n\in\mathbb{Z}. $$ The Fourier series of $f$ converges uniformly, as well as the series derived term by term. However, we only know that $f_m$ and $f$ are integrable. We could change their values on a finite set of points and the Fourier coefficients would be the same. We cannot conclude that the sum of the Fourier series is equal to $f$. Consider for instance $f$ he periodic extension of $f(x)=0$ if $0<x<2\,\pi$ and $f(0)==1$. All conditions are met, but $f$ is not continuous.
If we know that $f$ is continuous, it will be equal to the sum of the Fourier series, which is a $C^1$ function because of the decay of the Fourier coefficients.