# Limit of Series of Distributions

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.

• No big deal, but shouldn't it be $\langle \mathcal{F}\psi_k,\phi\rangle = \langle \psi_k,\mathcal{F}\phi\rangle$? – md2perpe Oct 19 '18 at 17:34

Hint: $$\sin^2(k\,x)=\frac{1-\cos(2\,k\,x)}{2}.$$
• I would then obtain that $\lim_{k\to\infty}\langle \psi^2_k,\phi\rangle = \| \phi\|_{L^1}$, is this correct? – EpsilonDelta Oct 19 '18 at 15:48
• No, the limit would be $$\frac12\int_{-\infty}^\infty\phi(x)\,dx.$$ – Julián Aguirre Oct 19 '18 at 15:50
• Why bother with the Fourier transform? By the way, the $\sin(k\,x)$ case is just the Riemann-Lebesgue lemma. – Julián Aguirre Oct 19 '18 at 15:55
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.