Let $(\Omega_i,\mathcal A_i,\mu_i)$ be a $\sigma$-finite measure space and $k\in L^2(\mu_1\otimes\mu_2)$. We know that $$Tf:=\int k(\;\cdot\;,x_2)f(x_2)\;\;\;\text{for }f\in L^2(\mu_2)$$ is a linear Hilbert-Schmidt operator from $L^2(\mu_2)$ to $L^2(\mu_1)$. In particular, $T$ is compact and hence $\sigma(T)\setminus\{0\}=\sigma_p(T)\setminus\{0\}$.
Can we choose $k$ such that
- $0\in\sigma_p(T)$;
- $0\in\sigma_c(T)$;
- $0\in\sigma_r(T)$?
I'm also interested in the same question, but with a more general (but still compact) integral operator and would be very happy if someone could provide a good reference on that.
To make life simple let us assume that both measure spaces coincide with ${\mathbb N}=\{1, 2, 3, \ldots \}$ under counting measure. Thus the integral kernel $$ k=(k_{i, j})_{i, j\in {\mathbb N}} $$ turns out to be the same thing as the matrix of the associated integral operator on $\ell^2$, relative to the standard orthonormal basis.
Choosing $$ k(i, j) = \left\{ \matrix{ \displaystyle\frac1{i}, & \text{if } i=j, \cr 0, & \text{otherwise}, }\right. $$ gives rise to the diagonal operator with matrix $$ \left[\matrix{ 1 & 0 & 0 & \cdots \cr 0 & \frac12 & 0 & \cdots \cr 0 & 0 & \frac13 & \cdots \cr \vdots & \vdots & \vdots & \ddots \cr }\right] $$ which is injective and has dense range, hence $0$ lies in the continuous spectrum of $T$. If we instead take $$ \left[\matrix{ 0 & 0 & 0 & \cdots \cr 1 & 0 & 0 & \cdots \cr 0 & \frac12 & 0 & \cdots \cr 0 & 0 & \frac13 & \cdots \cr \vdots & \vdots & \vdots & \ddots \cr }\right] $$ we get what is sometimes called a weighted shift, and which is again injective, but its range is not dense because it is orthogonal to the first basis vector. Consequently $0$ lies in the residual spectrum of $T$.
Finally, the zero operator has $0$ in its point spectrum.