Eigenvalues of second kind Fredholm integral operator

101 Views Asked by At

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.