Show that compact operator cannot have a bounded inverse on a infinite-dimensional Hilbert space Show that compact operator cannot have a bounded inverse on a infinite-dimensional Hilbert space.
Hint: use $A*A^{-1}=E$.
I've found a pretty useful topic here: https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space. Can someone explain it with $A*A^{-1}=E$.
Thank you.
 A: One of the properties of compact operators is that they form a two sided ideal in the algebra of bounded linear operators on $H$. If $T$ is compact and $S$ is any bounded linear operator (not necessary compact) then $TS$ and $ST$ are compact. 
So if we assume  that $A$ is compact and invertible then $I=AA^{-1}$ must be compact as well. But there is a theorem which states that the identity operator in an infinite dimensional Banach space cannot be compact. So we get a contradiction. 
A: Suppose that $A$ is compact and $A$ has bounded inverse $A^{-1}$. Let $B(0,1)$ be the closed ball of radius $1$; we know that $A(B(0,1)$ is contained in a compact $C$, this implies that $A^{-1}(C)$ which contains $B(0,1)$ is compact since the image of a compact by a continuous map is compact, this impact that the closed unit ball is compact contradiction since the closed unit ball in an infinite dimensional Hilbert space is not compact.
A: A more expository answer, even though both other answers are great:
If $T$ is a compact operator, then so is $TS$ for any bounded operator $S$. Taking $S=T^{-1}$ shows that the identity operator is compact. Therefore, every bounded sequence has a convergent subsequence. But we can use Riesz's lemma that says
"If $X$ is a Banach space and $Y$ is a closed, proper subspace of $X$, then for each $\varepsilon>0$ there exists $x_\varepsilon\in X$ of norm $1$ such that $\text{dist}(x,Y)\geq1-\varepsilon$"
To construct a bounded sequence that has no convergent subsequences as follows:
Take $x_1\in X$ a unit vector in $X$ and set $Y=\text{span}\{x_1\}$. Then $Y$ is closed and proper (finite dimensional), so for $\varepsilon=1/2$ we get a unit vector $x_2$ such that $\text{dist}(x_2,Y)\geq1/2$. Thus $\|x_1-x_2\|\geq1/2$. Now do the same for $Y=\text{span}\{x_1,x_2\}$ to obtain $x_3\in X$ such that $\|x_3-x_1\|\geq1/2, \|x_3-x_2\|\geq1/2$. In this way we get a sequence $(x_n)$ of unit vectors such that for all $n,m$ it is $\|x_n-x_m\|\geq1/2$. This has the desired property.
