Theorem: Let $A:U \subset X \rightarrow Y$ be a continuous and compact operator from a subspace $U$ of a normed space $X$ into a normed space $Y$. Then the equation of the first kind $A\phi=f$ is improperly-posed if $U$ is not of finite dimension.
Are there conditions on the operator $U$ that make the problem well-posed (A is bijective and $A^{-1}$ is continuous) for infinite-dimensional $U$?
What is the intuition behind the fact (only for compact operators or not?) that infinite-dimensional spaces couldn't construct well-posed problems?
If a compact operator $K:X\to Y$ is invertible, then $$I_X=K^{-1}K$$ is compact, since the composition of compact operators is compact. Similarly, $I_Y$ is compact. It follows that both $X,Y$ are finite-dimensional.