[Math] If it exists, the inverse of a compact linear operator in infinite dimensional space cannot be bouded

compact-operatorsfunctional-analysisinverse

I have been reading some posts on here that I think are related such as this and this. I am still having a tough time coming up with a nice proof for my question.

Question: If a compact linear operator $T:X \rightarrow X$ on an infinite dimensional normed space $X$ has an inverse which is defined on all of $X$, show that the inverse cannot be bounded.

I found this in $8.3.8$ of Erwin Kreyszig functional analysis.

Best Answer

Suppose $T^{-1}$ existed and was bounded. Then $T^{-1}(B_X) \subseteq KB_X$ for some $K > 0$. Taking $T$ on both sides yields $$B_X = T(T^{-1}(B_X)) \subseteq T(KB_X) = KT(B_X).$$ Since $\overline{T(B_X)}$ is compact, it follows that $B_X$ is a closed subset of a compact set, and hence is compact, which implies $X$ is finite-dimensional.

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