$g_n(t)=\int_0^1\cos^3 (t+nu)h(u)\;du$
I noticed that $\left|g_n(t)-g_n(t_0)\right|\le\sup_{u\in[0,1]}|h(u)|\int_0^1\left|\cos^3(t+nu)-\cos^3(t_0+nu)\right|\;du$
But how do I prove that this one can be arbitrarily small when $t\to t_0$?
$g_n(t)=\int_0^1\cos^3 (t+nu)h(u)\;du$
I noticed that $\left|g_n(t)-g_n(t_0)\right|\le\sup_{u\in[0,1]}|h(u)|\int_0^1\left|\cos^3(t+nu)-\cos^3(t_0+nu)\right|\;du$
But how do I prove that this one can be arbitrarily small when $t\to t_0$?
As $a^3-b^3=(a-b)(a^2+ab+b^2)$, we have $|g_n(t)-g_n(t_0)|\leq 3\lVert h\rVert_{\infty}\int_0^1|\cos(t+nu)-\cos(t_0+nu)|du.$ As $\cos$ is a $1$-Lipschitz continuous map, $|g_n(t)-g_n(t_0)|\leq 3\lVert h\rVert_{\infty}\|t-t_0|,$ and we are done.
Estimate the integrand using the mean value theorem.