Let $f: \mathbb{R} \to \mathbb{C}$ be a trigonometric polynomial with degree $N > 0$ and $g \in L^1(\mathbb{T})$. I'd like to show that $$\lim_{n \to \infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)g(nt)\ dt = \hat{f}(0)\hat{g}(0).$$
I have learned that, $$f(x) = \sum_{m = -N}^N\hat{f}(m)e^{imx}$$ and if $g_n(t):= g(nt)$, then \begin{equation}\hat{g}_n(m) = \begin{cases} \hat{g}(m/n),\ \ \ \text{if $n$ divides $m$}\\ 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{otherwise}. \end{cases}\end{equation} So, we have then \begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)g(nt)\ dt = \frac{1}{2\pi}\sum_{m = -N}^N\hat{f}(m)\int_{-\pi}^{\pi}g(nt)e^{imt}\ dt. \end{align*} I thought I should be able to use $\hat{g}_n(m)$ but I don't have $e^{-imt}$. Is there a trick that I haven't use? Putting $e^{-imt}e^{imt}$ inside the integral doesn't seem to help as well.
