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.

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

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

  • $\begingroup$ I would then obtain that $\lim_{k\to\infty}\langle \psi^2_k,\phi\rangle = \| \phi\|_{L^1}$, is this correct? $\endgroup$ – EpsilonDelta Oct 19 '18 at 15:48
  • $\begingroup$ No, the limit would be $$\frac12\int_{-\infty}^\infty\phi(x)\,dx.$$ $\endgroup$ – Julián Aguirre Oct 19 '18 at 15:50
  • $\begingroup$ Yes, of course, I missed out on the factor and the absolute... Do you know if my other solution is correct (the non squared one)? Also do you have any idea how to complete my other approach via FT? $\endgroup$ – EpsilonDelta Oct 19 '18 at 15:51
  • 1
    $\begingroup$ Why bother with the Fourier transform? By the way, the $\sin(k\,x)$ case is just the Riemann-Lebesgue lemma. $\endgroup$ – Julián Aguirre Oct 19 '18 at 15:55
  • $\begingroup$ Out of curiosity, does the RL lemma also hold for inverse FT? $\endgroup$ – EpsilonDelta Oct 19 '18 at 16:09

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.


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