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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.

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