Suppose we have a integral equation $$\int_{-1}^1 \frac{\text{sin }c(x-y)}{\pi (x-y)}\psi(y)dy=\lambda \psi(x),\quad|x|\le1.$$

By the Fredholm equation theory, we know that this equation has solutions in $L^2(-1,1)$ only for a discrete set of real positive values of $\lambda$, say $\lambda_0\ge\lambda_1\ge\lambda_2\ge...,$ and as $n\to \infty$, lim $\lambda_n=0.$ The corresponding solutions, or eigenfunctions, $\psi_0(x),\psi_1(x),\psi_2(x),...$ can be chosen to be real and orthogonal on $(-1,1).$

The author extended the range of definition of the $\psi$'s in this way: $$(*)\qquad\qquad\qquad\qquad \psi_n(x)=\frac{1}{\lambda_n}\int_{-1}^1 \frac{\text{sin }c(x-y)}{\pi (x-y)}\psi_n(y)dy,\quad|x|\gt1.$$

And he said that, by a simple calculation, we can show that these eigenfunctions are orthogonal on $(-\infty,\infty)$ as well as on $(-1,1)$ as already noted. We normalize them to have unit energy, so that$$\int_{-\infty}^{\infty} \psi_{n}(x) \psi_{m}(x) d x=\delta_{m m}$$ and it then follows that $$\int_{-1}^{1} \psi_{n}(x) \psi_{m}(x) d x=\lambda_{n} \delta_{m n}.$$

My question is that, how to show the orthogonality on $(-\infty,\infty)\,?$ And how to derive the last equality from $\int_{-\infty}^{\infty} \psi_{n}(x) \psi_{m}(x) d x=\delta_{m m}\,?$

I tried to calculate the inner product of $\psi_n$ and $\psi_m$, and then plug in $(*)$, and used Fubini's theorem. But the triple integral is very messy and I can't really calculate it. And for the derivation problem, I have no idea.

Any help will be appreciated.


He hid one important piece of information from you, namely that $\frac{\sin cx}{x}$ is the Fourier transform of the characteristic function of $[-c,c]$ (up to a normalizing factor, which depends on where you want to place $2\pi$ in the definitions). So, if we denote by $P$ the projection from $L^2(\mathbb R)$ to $L^2([-1,1])$, i.e., $Pf=f\chi_{[-1,1]}$ and by $Q$ the similar projection on the Fourier side: $\widehat{(Qf)}=\widehat f\chi_{[-c,c]}$, the original $\psi_m$ (extended by $0$ outside $[-1,1]$ are the eigenfunctions of the compact self-adjoint operator $PQP$, while the new $\psi_m$ (let's call them $\widetilde\psi_m$ to avoid confusion) are just $\widetilde\psi_m=\lambda_m^{-1}QP\psi_m$ without the final projection to $L^2([-1,1])$.

Now we trivially get (in $L^2(\mathbb R)$) $$ \langle\widetilde \psi_n,\widetilde\psi_m\rangle=\lambda_n^{-1}\lambda_m^{-1}\langle QP\psi_n,QP\psi_m\rangle=\lambda_n^{-1}\lambda_m^{-1}\langle PQP\psi_n,\psi_m\rangle= \lambda_m^{-1}\langle\psi_n,\psi_m\rangle $$ The rest should be clear.

  • $\begingroup$ Hi fedja, thanks for your post. I still got two problems in understanding your post. Q1: How did you derive $\widetilde\psi_m=\lambda_m^{-1}QP\psi_m$ from $\lambda_m\psi_m=PQP\psi_m?$ Q2: How did you obtain the equality $\langle QP\psi_n,QP\psi_m\rangle =\langle PQP\psi_n,\psi_m\rangle?$ I know $P^*=P$, but I don't know how to deal with $Q^*.$ Thanks. $\endgroup$ – Sam Wong May 28 '19 at 15:45
  • $\begingroup$ @SamWong Q1 P does nothing to $\psi_m$ and $Q$ is the convolution with $\frac{\sin cx}x$ on the space side. Q2 $Q^*=Q$ and also $Q^2=Q$. $\endgroup$ – fedja May 28 '19 at 21:00
  • $\begingroup$ Hi fedja, I think $\frac{sin\,cx}{x}$ should be the $\color{red}{Inverse}$ Fourier transform of the characteristic function of $[-c,c]$. And now it is clear that $Q$ is the convolution with $\frac{sin\,cx}{x}$. Thanks for your post. $\endgroup$ – Sam Wong Jun 3 '19 at 3:59

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