Let $K\in L^2((0,1)\times(0,1))$ and consider the operator defined in $L^2(0,1)$ by $$Lu(x):=u(x)-\int_0^1K(s,x)u(s)ds.$$ What kind of assumption might I impose on $K$ such that $\lambda=1$ will be not an eigenvalue of the operator $L$?. Any ideas?. Thank you.
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If $\lambda=1$ is an eigenvalue, then $\int_0^1 K(s,x)u(s) ds=0$ for the corresponding eigenfunction. Hence a condition to rule this out is that $\{K(s,x),x \in (0,1)\}$ spans a dense subspace of $L^2$.
answered Mar 23, 2020 at 16:52
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$\begingroup$ Thank you Mr Renardy for the answer. I'm wondering if there exists such a kernel. For instance, the constant kernel doesn't fulfill this property. Do you have an example in mind. Thank you Mr Renardy again. $\endgroup$– GustaveCommented Mar 24, 2020 at 17:22