# An Orthogonality Problem of Eigenfunctions of homogeneous Fredholm equation

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.
• 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. – Sam Wong May 28 '19 at 15:45
• @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 $Q^2=Q$. – fedja May 28 '19 at 21:00
• 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. – Sam Wong Jun 3 '19 at 3:59