The following is an exercise on Pg. $110$ in Rajendra Bhatia's Notes on Functional Analysis. $\mathcal B(X)$ is the set of all bounded linear operators on $X$, an infinite-dimensional Banach space.
Prove that the set of right-invertible operators is not dense in $\mathcal B(X)$. Similarly, show that the set of left-invertible operators is also not dense in $\mathcal B(X)$.
Notation: Let
- $\mathcal G$ or $\mathcal G(X)$ be the set of all invertible operators in $\mathcal B(X)$.
- $\mathcal G_R$ or $\mathcal G_R(X)$ be the set of all right-invertible operators in $\mathcal B(X)$.
- $\mathcal G_L$ or $\mathcal G_L(X)$ be the set of all left-invertible operators in $\mathcal B(X)$.
Certainly, $\mathcal G\subset \mathcal G_R$ and $\mathcal G\subset \mathcal G_L$. If $X$ is finite-dimensional, I have shown that $\mathcal G$ is dense in $\mathcal B(X)$. If $X$ is infinite-dimensional, then this is not necessarily the case. This can be seen by considering $X = \ell^2$, as in this post. The exercise above asks us to show that $\mathcal G_R(X)$ and $\mathcal G_L(X)$ are not dense in $\mathcal B(X)$. I am confused here, do we have to show this for every infinite-dimensional $X$, or produce a particular $X$ as a counterexample? Any hints would be appreciated.
Best Answer
You can show it for all $X$, but like most things in Hilbert space theory, it's easy to generalize the proof from $\ell^2$ to arbitrary $X$
The key point is that the idea in linked answer can be distilled in the following: Assume there is $A\in B(H)$ such that $A$ is left-invertible and not right-invertible. Then the ball of radius $1/\|A_L\|$ around $A$ is disjoint from the set of right-invertible operators. The proof is the same. Let $A_L$ denote the left-inverse of $A$, and assume that $B$ is right-invertible, with right-inverse $B_R$, and $\|B-A\|<1/\|A_L\|$. Then $$\|I-A_LB\|=\|A_L(A-B)\|\le \|A_L\|\|A-B\|<1.$$ Thus $A_LB$ is invertible, and in particular, there is $C$ so that $CA_LB=I$, but then $BCA_L=I$, so $A_L$ is left-invertible, and thus $A$ is right-invertible, a contradiction.
A similar statement holds with the chiralities switched.
Thus your question is reduced to showing that there are left-invertible, but not right-invertible elements, and vice-versa. For this, just take a countable orthonormal sequence $e_1,\dots e_n$. Define $Ae_i=e_{i+1}$, and $Ay=y$ for any $y$ orthogonal to all $\{e_n\}_n$. One can easily check this has the same properties as right-shift on $\ell^2$. Similarly one can define a left-shift operator.