Fourier series: Show that $f$ is a trigonometric polynomial

Question

Let $N\in\mathbb{N}$ and $f_m:\mathbb{R}\to\mathbb{R}$, continuous functions and periodic, $T=2\pi$. Let's assume that $f_m \to f$ uniformly and for all $m\ge 1$: $$\left| \hat{f_m}(n)\right| \le \frac{\left|n\right|}{m}$$

Prove that $f$ is a trigonometric polynomial.

So I started with couple of things in mind. For example, writing $f$ as a Fourier series, but I wasn't able to get something significant out of it.

I'd be glad for a solid lead.

Thanks.

Answer 1

For all $n\in\mathbb{Z}$, by uniform convergence: $$ \hat f(n)=\int_0^{2\pi}e^{inx}\,f(x)\,dx=\lim_{m\to\infty}\int_0^{2\pi}e^{inx}\,f_m(x)\,dx=\lim_{m\to\infty}\hat{f_m}(n). $$