# A variational problem involving integral

Let $$W\gt 0,\,T\gt 0$$ be fixed.

Let $$R(f)$$ be an arbitraty function in $$L^2(-W,W).$$

Define $$\alpha_{R}:=\frac{\int_{-W}^W\,df^\prime\int_{-W}^W\,df^{\prime\prime}\frac{sin\,\pi T(f^\prime-f^{\prime\prime})}{\pi (f^\prime-f^{\prime\prime})}R(f^{\prime\prime})\overline{R(f^\prime)}}{\int_{-W}^W\,df^\prime R(f^\prime)\overline{R(f^\prime)}}$$, where $$\overline{R(f^\prime)}$$ means the complex conjugate of $$R(f^\prime)$$.

The writter said that, any $$\overset{\sim}{R}$$ such that $$\alpha_{\overset{\sim}{R}}=\underset{R\in L^2(-W,W)}{max}a_R$$ must satisfy the integral equation $$\int_{-W}^W\,\frac{sin\,\pi T(f^\prime-f^{\prime\prime})}{\pi (f^\prime-f^{\prime\prime})}{\overset{\sim}{R}}(f^{\prime\prime})df^{\prime\prime}=\alpha_{\overset{\sim}{R}}\,{\overset{\sim}{R}}(f^\prime),\qquad|f^\prime|\le W.$$

My problem is that why satisfying this integral equation is a necessary condition for a maximizer $${\overset{\sim}{R}}$$?

Sorry for the ugly notations, which are from a article written in the $$80's$$.

My attempt: I try to multiply the denominator to both sides of the first equation and then substract the left hand side from both sides. And then apply the Cauchy-Schwarzt inequality. I did obtain a similar equality like the second equation. But in that process, it was maximizing the difference rather than $$\alpha_R$$ itself. I have no idea now and kind of frustrated.

Any help will be greatly appreciated.