Let $f$ be a function on the circle. For each $N \geq 1$ the discrete Fourier coefficients of $f$ are defined by $$a_N(n)=\frac{1}{N}\sum_{k=1}^Nf(e^{2 \pi i k/N})e^{-2 \pi i kn/N}\ \ \ \text{for}\ n \in \mathbb{Z}$$ we also let $$a_n=\int_0^1 f(e^{2 \pi i x})e^{-2 \pi i nx} dx$$ denotes the ordinary Fourier coefficients of $f$. Show that if $f$ is continuous on the circle, then $a_N(n)\rightarrow a_n$ as $N \rightarrow \infty$.
Since $f$ is continuous, it is bounded on the circle, and hence it is bounded since the circle is compact.
But how to guarantee that $f(e^{2 \pi i k/N})$ is bounded as well?
I would appreciate any help with this
Thank you in advance.
Hint: $g_n(x)=f(e^{2i\pi nx})e^{-2i\pi nx}$ is continuous, so it is Riemann-integrable. Now notice that $a_N(n)$ is a Riemann sum.