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$$\sqrt{2}f(x) =\lim_{\delta \to 0^{+}}\left[x-i\delta-\int_{-1}^{1} \frac{|f(y)|^2}{y-i\delta-x}dy\right]$$

I'd like to know if there is a solution for $f\colon(-1,1) \to\mathbb{C}$. Of course if it doesn't cause any problems, the interval may also be the closed one $[-1,1]$.

I know,that if the $|f(y)|^2$ was replaced by $f(y)$ that would be a Fredholm-equation.

P.S.: I'm am new to stack exchange, so please tell me, if you'd rather deem this to be a question for mathoverflow.

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While I normally don't change $]a,b[$ notation to $(a,b)$ notation for open intervals, I found the change here greatly improving the readability. –  Asaf Karagila Dec 14 '12 at 14:21
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1 Answer

Note: this is not a full solution, but it is possible to reduce the integral equation to an equation for a single real function. Maybe somebody else knows how to solve the final integral equation...

The integral equation is nonlinear that is why (as far as I know) it does not have a name. Progress towards its solution be made by remembering Sokhotsky's Formula $$\lim_{\delta\to 0^+} \frac{1}{z-i \delta} = i \pi \delta(z) + \mathcal{P} \frac{1}{z}.$$

Applying it to your equation while assuming $|x| <1$, we obtain $$\sqrt{2} f(x) = x - \int_{-1}^1 |f(y)|^2 \left[i\pi \delta(y-x) + \mathcal{P} \frac{1}{y-x}\right] dy = x- i \pi |f(x)|^2 - \int_{-1}^1 \mathcal{P} \frac{|f(y)|^2 dy}{y-x}. $$

Introducing $f(x) = u(x) + i v(x)$, we can take the imaginary part of the equation $$ \sqrt{2} v(x) = -\pi [u(x)^2 + v(x)^2] \qquad(1)$$ and the real part reads $$ \sqrt{2} u(x)= x- \int_{-1}^1 \mathcal{P} \frac{[u(y)^2+v(y)^2] dy}{y-x}.\qquad(2)$$

From equation (1) and the fact that $u$ is real, we obtain $-\sqrt{2}/\pi\leq v\leq 0$. This suggests the change of variable $v=-1/\sqrt{2}\pi + \delta v$ after which the first equation assumes the simple form $$ 2 \pi^2 (u^2 + \delta v^2) = 1.$$ This constraint can easily be satisfied by introducing $\theta(x)$ such that $$ u= \frac1{\sqrt{2} \pi} \cos\theta \qquad \text{and}\qquad \delta v= v+\frac{1}{\sqrt{2}\pi} =\frac1{\sqrt{2} \pi} \sin \theta.$$

Plugging this expression in equation (2) yields $$ \cos \theta(x) = \pi x + \int_{-1}^1\mathcal{P} \frac{\sqrt{2} v(y)dy}{y-x} = \pi x + \frac1\pi \log\left| \frac{1+x}{1-x} \right|+ \frac{1}{\pi}\int_{-1}^1\mathcal{P} \frac{\sin\theta(y)dy}{y-x} .$$

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