Given $X,Y$ Banach spaces and $T \in L(X,Y)$, show that the following sentences are equivalent:

A) there exists $S \in L(Y,X)$ such that $S(T(x))=x$ for all $x \in X$.

B) $T$ is injective and $T(X)$ is a complemented space of $Y$.


I was given this exercise in my functional analysis course but I don’t know how to solve this.

All I have understood so far in this exercise are the following:

  • I was given the following definition of "complemented space'': a closed subspace $M$ is complemented in $N$ if exists a topological complement of $M$ in $N$ or equivalently if there exists a linear continuous projection $P$ in $N$ such that $𝑃(N)=M$;
  • $L(X,Y)$ means the set of all continuous linear operators from $X$ to $Y$.
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    $\begingroup$ I have undeleted this question. I did this mainly because I find the answer to be clear and good, and the question is reasonable. I recall coming across these sorts of questions with some confusion when I was learning functional analysis. $\endgroup$ – davidlowryduda Jan 23 '19 at 21:14

$(B)\Rightarrow (A)$. Since $T(X)$ is topologically complemented, let $P:Y\to T(X)$ be a bounded linear projection. Then the operator $$P \circ T: X\to T(X)$$ is:

  1. bounded, because both $P$ and $T$ are bounded;
  2. injective, because $T$ is injective and the restriction of $P$ to $T(X)$ is the identity;
  3. surjective, again because the restriction of $P$ to $T(X)$ is the identity.

Moreover, $T(X)$ is closed, so it is a Banach space. Hence by the inverse mapping theorem, $P\circ T$ has a bounded linear inverse $(P\circ T)^{-1}:T(X)\to X$. We now define an operator $S:Y\to X$ by $$S:=(P\circ T)^{-1}P $$ then $S$ is bounded, and $S\circ T=(P\circ T)^{-1}P\circ T=I$ so that $STx=x$ for all $x\in X$ as required.

$(A)\Rightarrow (B)$. Suppose that $Tx=Ty$. Then $x=STx=STy=y$. So $T$ is injective. Since $S$ is bounded, we have $$\|x\|=\|STx\|\leq \|S\|\|Tx\| $$ Thus $$\|Tx\|\geq \|S\|^{-1}\|x\|\qquad \forall x\in X $$ i.e. $T$ is also bounded from below (other than from above), and hence it is an isomorphism between $X$ and its range $T(X)$. In particular $T(X)$ is a Banach space and therefore it is closed.

It remains to find a continuous linear projection from $Y$ to $T(X)$. Define $$P:=TS $$ Then

  1. $P$ is bounded, because $T$ and $S$ are bounded;
  2. the range of $P$ is contained in $T(X)$;
  3. for all $y=Tx\in T(X)$, we have $$Py=P(Tx)=T(STx)=Tx $$ so that the restriction of $P$ to $T(X)$ is the identity. This completes the proof.
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  • $\begingroup$ In the first part, when you define the inverse, it is $ (P \circ T)^{-1}: T(X) \rightarrow X$ , not $T(X) \rightarrow T$, right? $\endgroup$ – Maggie94 Jan 8 '19 at 19:43
  • $\begingroup$ In the second part, why we get that $T$ is bounded if $\Vert T(x) \Vert \geq \Vert S^{-1} \Vert \Vert x \Vert $ ? $\endgroup$ – Maggie94 Jan 8 '19 at 19:48
  • $\begingroup$ We don't, $T$ is bounded by assumption. However, it allows us to conclude that it is bounded from below. An operator that is both bounded and bounded from below is an isomorphism onto its range. This is rather easy to prove, but you could also prove that $T(X)$ is closed directly from the relation $\|Tx\|\geq \|S\|^{-1}\|x\|$. $\endgroup$ – Lorenzo Quarisa Jan 8 '19 at 20:02
  • $\begingroup$ Why I can get that $T(X)$ is closed from the last inequality yoy wrote? $\endgroup$ – Maggie94 Jan 8 '19 at 20:22
  • $\begingroup$ Take a Cauchy sequence in $T(X)$, using that inequality and the continuity of $T$ you show that it converges to an element of $T(X)$. $\endgroup$ – Lorenzo Quarisa Jan 8 '19 at 20:59

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