Limit of Series of Distributions

Question

Let $\displaystyle \langle \psi_k,\phi\rangle = \int_{-\infty}^\infty \sin(kx)\phi(x)\,\mathrm{d}x,\ \phi \in \mathcal{C}_c^\infty(\Omega),\Omega\subset \mathbb{R}$ open. That means $\phi$ is a testfunction.

I want to compute

$\displaystyle \lim_{k\to \infty}\langle \psi_k,\phi\rangle$ and also $\displaystyle \lim_{k\to \infty}\langle \psi^2_k,\phi\rangle$

I know that for the fouriertransform $\mathcal{F}$ as a unitary operator it holds true that

$\langle \mathcal{F}\psi_k,\phi\rangle = \langle \psi_k,\mathcal{F}^{-1}\phi\rangle$

Using this I compute the inverse fourier transform of $\sin(kx)$ which yields

$\displaystyle i\sqrt{\frac{\pi}{2}}\delta(k+x)-i\sqrt{\frac{\pi}{2}}\delta(k-x)$

then I get

$\displaystyle \lim_{k\to\infty}\int_{-\infty}^{\infty}i\sqrt{\frac{\pi}{2}}\mathcal{F}^{-1}\phi(x) -i\sqrt{\frac{\pi}{2}}\mathcal{F}^{-1}\phi(-x)\,\mathrm{d}x$

which does not really help. Although I could solve the first (non-squared) one by using integration by parts (limit is then zero), I could not solve the squared version of it by using this technique, so I tried another strategy, which is the one I showed above.

Answer 1

Hint: $$ \sin^2(k\,x)=\frac{1-\cos(2\,k\,x)}{2}. $$

Answer 2

Your method will work fine; $(\hat \psi, \phi) = (\psi, \hat \phi)$ gives $$(\sin^2 k x, \phi) = \frac 1 4 (2 \delta(w) - \delta(w - 2 k) - \delta(w + 2 k), \hat \phi) \to \frac {\hat \phi(0)} 2 = \left( \frac 1 2, \phi \right).$$ The other two delta terms tend to zero because $\hat \phi$ is a function of rapid decay.