Let's take the operator $L=(\partial_x^2+m^2)\delta(x-y)$ as an example. It operates in the function space in the following way $$\int dy\ (\partial_x^2+m^2)\delta(x-y)f(y)=(\partial_x^2+m^2)f(x).$$ The eigen equations are $$\int dy\ (\partial_x^2+m^2)\delta(x-y)f(y)=(\partial_x^2+m^2)f(x)=k(\sigma)f(x,\sigma),$$ where $\sigma$ simply labels the eigenfunctions. For our example we have $k(\sigma)=-\sigma^2+m^2$ and $f(x,\sigma)=e^{i\sigma x}$. And for its inverse, we have $$\int dy\ (\partial_x^2+m^2)\delta(x-y)G(y-z)=\delta(x-z),$$ namely $$(\partial_x^2+m^2)G(x-z)=\delta(x-z).\tag{1}$$ $G(x-z)$ is simply the Green's function. All these stuff are completely analogy with the discrete case. The trace is defined as $$tr L=\int d\sigma k(\sigma)$$ and we can obtain the determinant by $\log \det L=tr \log L$.
Now comes my question, since from Eq.(1) we can not obtain an unique Green's function (recall the boundary conditions if we want to specify the solution). Then when a operator $L$ has no unique inverse, shouldn't it have vanishing determinant?