Suppose $\lambda\not=0\in\mathbb{C}$. Does the following system have a non trivial solution in $L^2 [0,1]$?
\begin{array} {lcl} \int\_0 ^1 f(y)\log|x-y|dy=\lambda f(x) \\\int\_0 ^1f(x)dx=0& \end{array} 



Solvability of a Fredholm system in $L^2$
I will give a partial answer. If you don't like it I can delete it. I might also be restating what fedja is saying, in which case I apologize. It isn't clear to me that this is what he/she had in mind though.


For $|\lambda| > 1$ the answer is no. We can implement a fixed point argument to show this.


First, let $X = \{f\in L^2([0,1] \; : \; f(1-x) = -f(x)\}$ (as fedja suggests). This is a vector space. Let $$T\_\lambda f(x) = \frac{1}{\lambda} \int\_0^1 f(y) log|x-y| dy$$ As fedja suggests, $T\_\lambda$ maps $X$ into $X$. Using Young's inequality one gets $$\left\| T\_\lambda f \right\|\_{L^2([0,1])} \leq \frac{1}{|\lambda|} \left\| f \right\|\_{L^2([0,1])} \left\| log(x) \right\|\_{L^1([0,1])} = \frac{1}{|\lambda|} \left\| f \right\|\_{L^2([0,1])}$$ So, $T\_\lambda$ is a contraction mapping from $X$ into $X$. Using the Banach fixed point theorem you get that there exists unique fixed point. Since 0 is a fixed point then that's the only one there is.


Not sure what to do if $|\lambda| \leq 1$.


