On the existence of a specific linear operator

Question

Say $X$ and $Y$ are two Banach spaces (of finite or infinite dimension), $T: X \rightarrow Y$ a bounded linear operator and suppose $T$ is injective. Then there exists $S: Y \rightarrow X$, linear and bounded, such that $ST = Id$.

I started off by observing that $T: X \rightarrow T(X)$ is surjective. Then $T: X \rightarrow T(X)$ is bijective and therefore invertible: there exists $S \in L(T(X), X)$ such that $ST = TS = Id$. May I conclude by extending $S$ by $0$ on $Y \setminus T(X)$? Is there another way of proving this result?

Answer 1

This is false. For example, take $T\colon \ell_\infty \to c_0$ given by $T(\xi_k)_{k=1}^\infty = (\xi_k / k)_{k=1}^\infty$. It is injective but it does not have a left-inverse as it is compact.

Even when $T$ is injective and has closed range (which is equivalent to the existence of $\delta>0$ such that $\|Tx\|\geqslant \delta\|x\|$ for all $x$) this need not hold. Indeed, let $T\colon c_0\to \ell_\infty$ be the inclusion map. There is no surjective operator $S\colon \ell_\infty\to c_0$ because $\ell_\infty$ is a Grothendieck space.

Answer 2

As written, your proof will generally not produce a linear operator. Consider $T:\ell^1\to\ell^1$ given by $$Tx(n)=x(n-1).$$ Then we have the left inverse $S:T(\ell^1)\to\ell^1$ defined by $$Sx(n)=x(n+1).$$ But, we have $y=\left(1,1,\frac{1}{4},\ldots,\frac{1}{(n-1)^2},\ldots\right)\in \ell^1\setminus T(\ell^1),$ and if we extend $S$ to $\tilde{S}$ by means of your proof, we have $$\tilde{S}y=0\neq \tilde{S}\left(1,0,0,\ldots\right)+\tilde{S}\left(0,1,\frac{1}{4},\ldots,\frac{1}{(n-1)^2},\ldots\right).$$

Generally, what is required is that $T(X)$ be a complemented subspace of $Y$. Because then $Y=T(X)\oplus Z$ for some closed subspace $Z$ of $Y$, and we can extend $S$ by linear extension of \begin{align} \tilde{S}y=\left\{ \begin{array}{lcl} Sy&:&y\in T(X)\\ 0&:&y\in Z \end{array} \right. \end{align} In this case, we also get preservation of norm, i.e. $\|\tilde{S}\|=\|S\|$.