I have to prove that the following limit goes to $0$:
$$\lim_{n\to\infty} \int_{-\infty}^{\infty} \left|u(\omega)\bigg(\frac{2n^2}{\omega^2}(1-\cos(\omega/n))-1\bigg)\right|^2dw$$
where $u \in L^2(\mathbb{R})$.
I want to move the limit inside to that I can substitute:
$$\cos(1/t) = 1 - \frac{1}{2t^2} + \frac{1}{4!t^4} - \text{...} $$
And the proof is completed.
I was thinking about using the dominated convergence theorem, but I am not sure how to use it. As a dominating sequence I would use that:
$$1-\cos(t) \leq t^2/2 $$
But then, what else should I make sure of?
We can use your inequality to show that $$ 0 \leq \frac{2(1-\cos(t))}{t^2} \leq 1$$ holds for $t \in \mathbb{R}$ .
If we define $f_n \colon \mathbb{R} \to \mathbb{R}$ to be your integrand, i.e. $$ f_n(\omega) = |u(\omega)|^2 \left[1 - \frac{2 (1-\cos(\omega /n))}{(\omega/n)^2}\right]^2 \, , $$ we immediately obtain $|f_n(\omega)| \leq |u(\omega)|^2$ for all $\omega \in \mathbb{R}$ and every $n \in \mathbb{N}$ from the above inequality.
We take $|u|^2$ as the dominating function from the theorem. It is integrable, as we have $u \in L^2 (\mathbb{R})$ . Therefore the dominated convergence theorem allows us to interchange the limit and the integration. But now we are done, since $(f_n)_{n\in \mathbb{N}}$ converges to zero pointwise almost everywhere (as mentioned in your question).