Define the operator $K$ on $L^2[0,1]$ by $(Ku)(x) = \int_0^1 k(x, y) u(y) dy$ where $k$ is a real-valued bounded continuous function with $k(x, y) > 0$. Consider the problem $Ku = f$. Let $X$ be the set of all $f \in L^2[0, 1]$ such that a unique solution $u = K^{-1} f$ exists. can we be assured that there is a constant $C$ such that $\|K^{-1}f\| \leq C\|f\|$ for all $f \in X$?
If $X$ is complete the answer is yes ($K$ is not only bounded but compact; apply the bounded inverse theorem). Suppose $X$ is not complete. I suspect the answer is no but I cannot think of any example that would show that $K^{-1}$ is not bounded. Any suggestions?