Consider the operator $$K:L^2(0,1)\rightarrow L^2(0,1) \\ u\rightarrow\int_0^1k(s,x)u(s)ds.$$ with $k\in L^2((0,1)\times(0,1)).$
I want to know under what assumption the kernel is reduced to zero. i.e. $ker(K)={0}$. I can say that if $k$ is a Green function for some differential operator this will be true. But what about the general case? Can we obtain a criteria for the injectivity by some expansion process on the $L^2$ basis?. Thank you.
Your operator $K$ is a Hilbert-Schmidt operator since its kernel belongs to $L^2$. As a result this is a compact operator whose spectrum contains a sequence of eigenvalues $\\{\lambda_k\not=0\\}$ with finite multiplicities such that $\lim_k\vert \lambda_k\vert=0$. To deal with the self-adjoint case, you can find an orthonormal set $\\{\mathbf e_k\\}$ such that $$ K\mathbf e_k=\lambda_k\mathbf e_k \quad \mathbb R\ni\lambda_k, \quad (\lambda_k)\in \ell^2 .$$ As a result, 0 will always belong to the spectrum even if $\\{\mathbf e_k\\}$ is an orthonormal basis, but $K$ will be injective in that case. Setting $$ E=\overline{\text{span}\\{\mathbf e_k\\}},$$ you get $ \ker K=E^\perp. $
