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}
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2$\begingroup$ I wrote some nonsense instead of what I really meant, sorry for that. The comment should read "The space of "odd" functions ($f(1-x)=-f(x)$) is invariant and the integral operator with the kernel $\log|x-y|$ is compact (Hilbert-Schmidt) and not identically zero there". $\endgroup$– fedjaCommented May 18, 2014 at 2:14
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$\begingroup$ Actually I wasnt able to see your previous comment.but this comment right up here totally makes sense.thanks $\endgroup$– BigMCommented May 18, 2014 at 5:54
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$\begingroup$ @fedja : So the answer is yes : all eigenfunctions with non zero eigenvalue of the integral operator restricted to the subspace of odd functions are solutions. $\endgroup$– jjcaleCommented May 18, 2014 at 19:25
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1 Answer
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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 $|\lambda| \leq 1$.
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$\begingroup$ How do we know that fixed point is not identically zero? $\endgroup$– BigMCommented May 18, 2014 at 6:29
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2$\begingroup$ I think $f(1-x)=-f(x)$ should hold almost everywhere as $L^2$-functions have no pointwise values. This answer actually shows that there are no nonzero solutions for $|\lambda|>1$. $\endgroup$ Commented May 18, 2014 at 9:03
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$\begingroup$ @Joni Ahhh yes, thanks for pointing out that there are no nontrivial solutions. You are correct. I apparently can't do math before I go to bed. $\endgroup$– k3thompsCommented May 18, 2014 at 12:03
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$\begingroup$ Say we know that there exists a non trivial solution for the first equation.can we find a non trivial function that satisfies both equations? $\endgroup$– BigMCommented May 18, 2014 at 15:04
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1$\begingroup$ @user48438 Interesting! I'll give the other case a bit more thought and I'll let you know if I think of anything. $\endgroup$– k3thompsCommented May 18, 2014 at 22:03