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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?

freishahiri
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  • Out of interest, how do you know that $T(X)$ is complete? I don't know how to prove this and am wondering if you could explain it to me. – Kenny Wong Feb 28 '17 at 20:39
  • I don't know if $T(X)$ being complete intervenes in this proof. I think that one needs $T(X)$ to be closed. The open mapping theorem may yield this result but I am not sure. –  Feb 28 '17 at 20:45
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    Well $T(X)$ being closed is the same as $T(X)$ being complete, isn't it, because $T(X)$ is a subspace of $Y$ which is complete? – Kenny Wong Feb 28 '17 at 20:46
  • Don't you need to check that $T(X)$ is closed/complete before you can even begin applying the open mapping theorem to get your $S : T(X) \to X$? Or am I missing something? – Kenny Wong Feb 28 '17 at 20:47
  • Indeed, there is an equivalence between the two. And you're right again, one needs $T(X)$ to be Banach in order to use the open mapping theorem. So I don't know how to prove that $T(X)$ is closed, sorry. –  Feb 28 '17 at 20:48
  • Let $T : l^1 \to l^1 $ map $(x_1, x_2, x_3, \dots) \mapsto (x_2, x_3, x_4 , \dots)$. What $S : l^1 \to l^1 $ satisfies $ST = id$? – Kenny Wong Feb 28 '17 at 20:57
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    @Kenny Your proposed $T$ is not injective. – Theoretical Economist Feb 28 '17 at 21:00

2 Answers2

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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. $\|\tilde{S}\|=\|S\|$.

Aweygan
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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.

Community
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Tomasz Kania
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