# integral convergence of function $e^{inx} f(x)$

Show that $n\to \infty \int_0^1e^{inx} f(x)dx=0$ for any continuous function on [0,1].

I tried to show the collection of {$e^{nix}:n\ge0$ } is algebra, separates points and do not vanishes at origin then I started using stone-weierstrass but lost in the middle. Please provide me some hints. Thanks in Advance

Hint: For step functions (those of the form $f=c_1f_1+\cdots+c_mf_m$ where each $f_j$ is the indicator function of an interval $[a_j,b_j]$) this is easy, since $$\int_0^1e^{inx}f(x)dx=\sum_{j=1}^m\int_0^1e^{inx}c_jf_j(x)dx=\sum_{j=1}^mc_j\int_{a_j}^{b_j}e^{inx}dx=\sum_{j=1}^m\frac{c_j}{in}(e^{inb_j}-e^{ina_j})$$ which can easily be shown to go to $0$. Now approximate an arbitrary function by step functions.