I am trying to prove that $$\langle\tilde{s}(\omega), \omega^{2}\tilde{s}(\omega)\rangle = -\langle s(t), \partial_t^2 s(t)\rangle$$
Where $\tilde{s}(\omega)$ is defined as the Fourier transform of a signal $s(t)$ such that $$\tilde{s}\omega =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} s(t)e^{-i\omega t} \, dt$$
My attempt at an answer:
We already know that $$\mathcal{F}[\partial_t s(t)](\omega) = i\omega \mathcal{F}[s(t)](\omega)$$ and hence it follows that $$\mathcal{F}[\partial_t^2 s(t)](\omega) = -\omega^2 \mathcal{F}[s(t)](\omega)$$
so
$$\langle \tilde{s}(\omega), \omega^2 \tilde{s}(\omega)\rangle = \langle\mathcal{F}[s(t)], -\mathcal{F}[\partial_t^2 s(t)]\rangle$$
At some point I'm going to have to red rid of the fourier transforms - but how do I go about doing that? Is it better to work in full integral notation and to switch to inner products at the end of the proof?
The Plancherel theorem may help you. It just says $\langle\tilde f,\tilde g \rangle = \langle f,g\rangle$.