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If $T$ is a linear operator between vector spaces $X,Y$ and $T^\ast$ denotes the algebraic adjoint, i.e. $T^\ast:Y^\ast\to X^\ast,y^\ast\mapsto y^\ast\circ T$, it is easy to see that if $T$ is injective (surjective), then $T^\ast$ is surjective (injective). In particular, if $T$ is bijective, then $T^\ast$ is bijective.

Now I've read that if $X,Y$ are Banach spaces and $T':Y'\to X'$ denotes the topological adjoint, then $T$ is bijective if and only if $T'$ bijective. How can we show this and why is it important that we are dealing with Banach spaces and that we've replaced the algebraic with the topological duals?

And are we only able to show the equivalence or does it even hold that if $T'$ is injective (surjective), then $T$ is surjective (injective)?

0xbadf00d
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  • What is $X'$ exactly? – Evangelopoulos Foivos Jan 29 '21 at 07:01
  • @EvangelopoulosF. $X'$ is the topological dual space of $X$. – 0xbadf00d Jan 30 '21 at 18:43
  • https://math.stackexchange.com/questions/346560/show-t-is-invertible-if-t-is-invertible-where-t-in-bx-t-in-bx/346640#346640 .. for a longer answer to the bijectivity part of the question – daw Jan 30 '21 at 19:13
  • see also https://math.stackexchange.com/questions/346560/show-t-is-invertible-if-t-is-invertible-where-t-in-bx-t-in-bx/346640#346640 – daw Jan 30 '21 at 19:15
  • At the bare minimum, you would need some Hahn-Banach condition. Otherwise, if $X\neq 0$ is such that $X'=0$, then the zero map $X\to X$ has bijective dual, but it is neither injective nor surjective. – tomasz Jan 30 '21 at 19:30
  • And likewise, if $X$ is as above and $Y\leq X$ is nonzero finite-dimensional, then $Y\to X$ is injective, but $X'\to Y'$ is not surjective. Of course, it is always true that if $T$ is surjective, then $T'$ is injective. – tomasz Jan 30 '21 at 19:46
  • @daw Why did you delete your answer to the other question? I think your approach works: https://math.stackexchange.com/a/4011183/47771. – 0xbadf00d Feb 03 '21 at 16:03

1 Answers1

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The implication: ``$T$ surjective $\Rightarrow$ $T'$ or $T^*$ is injective'' is quite easy to prove:

Assume $T'f=T'g$. Then $(T'f)(x)=(T'g)(x)$ and $f(Tx)=g(Tx)$ for all $x\in X$. Since $T$ is surjective, $f=g$ follows. Same for $T^*$. Note that when dealing with continuous operators here, surjectivity of $T$ can be replaced by: image of $T$ is dense in $Y$.

The implication: ``$T$ injective $\Rightarrow$ $T'$ surjective'' is false in general: take $T$ to be an injective compact operator between infinite-dimensional Banach spaces. Then $T'$ is compact as well, and the range of $T'$ is not closed. I have no idea/intuition about the algebraic dual $T^*$ for this implication.

The implication: ``$T'$ injective $\Rightarrow$ $T$ surjective'' is false as well. Take $T$ compact with dense range, then $T'$ is injective, but $T$ has not closed range.

The implication ``$T$ bijective $\Rightarrow$ $T'$ or $T^*$ bijective'' on the other hand is true, and can be proven by showing that $(T')^{-1}=(T^{-1})'$ and $(T^*)^{-1}=(T^{-1})^*$.

The implication ``$T'$ bijective $\Rightarrow$ $T$ bijective'' is a consequence of the closed range theorem: By the closed range theorem, the range of $T$ is closed and the annihilator of the nullspace of $T'$, so $T$ is surjective. Injectivity of $T$ is clear. Here, $X,Y$ have to be Banach spaces.

daw
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  • (a) Maybe you got me wrong. It was already clear to me that the implication "$T$ is surjective $\Rightarrow$ $T^\ast$ is injective" holds. Do you claim that the converse is true as well? (b) Actually, I thought the implication "$T$ is injective $\Rightarrow$ $T^\ast$ is surjective" holds as well. In fact, if $y^\ast\in\mathcal N(T^\ast)$, then $$0=\langle x,T^\ast y^\ast\rangle=\langle Tx,y^\ast\rangle;;;\text{for all }x\in X$$ and hence $$\langle y,y^\ast\rangle=0;;;\text{for all }y\in\mathcal R(T).\tag1$$ If $T$ is surjective, then $\mathcal R(T)=Y$ and hence $(1)$ yields $y^\ast=0$. – 0xbadf00d Jan 30 '21 at 18:47
  • Am I missing something? (c) What I really wanted to know is whether the converse implications in (a) and (b) hold as well. And, if they don't hold, how can we show that, if $X,Y$ are Banach spaces, it holds "$T$ is bijective $\Leftrightarrow$ $T'$ is bijective" (note that the algebraic dual $T^\ast$ is replaced with the continuous dual $T'$). – 0xbadf00d Jan 30 '21 at 18:47
  • (a) I do not know (b) your proof is wrong: it proves $T$ surjective $\Rightarrow$ $T^*$ is injective. The converse is wrong at least for continuous duals. One can at best prove that the image of $T$ is dense, but it might not be closed. – daw Jan 30 '21 at 19:08
  • (c) is a consequence of the closed graph theorem – daw Jan 30 '21 at 19:12
  • (b) Sorry, I've mistakenly provided the proof of the wrong direction. The correct one goes as follows: Assume $A$ is injective. Let $x^\ast\in X^\ast$. Since $A$ is injective, $$\langle y,y^\ast\rangle:=\langle A^{-1}y,x^\ast\rangle;;;\text{for }y\in\mathcal R(A)$$ is well-defined. By construction, $$\langle x,A^\ast y^\ast\rangle=\langle Ax,y^\ast\rangle=\langle A^{-1}Ax,x^\ast\rangle=\langle x,x^\ast\rangle\tag2$$ for all $x\in X$; i.e. $A^\ast y^\ast=x^\ast$. – 0xbadf00d Jan 31 '21 at 05:42
  • At which stage does this proof fail when the algebraic duals are replaced by the topological ones? Shouldn't the same proof be valid? – 0xbadf00d Jan 31 '21 at 05:43
  • $y^*$ might be unbounded (if, e.g., $A$ is compact) – daw Jan 31 '21 at 10:22
  • (b) You're right. However, the proof should remain to be correct, if we assume that $X,Y$ are complete and $\mathcal R(T)$ is closed, since then $T^{-1}:\mathcal R(T)\to X$ is bounded by the bounded inverse theorem. Do you agree or am I missing something? In any case, I've read that the claim holds without these assumptions. So, maybe only my proof is not working. But how do we prove it then? – 0xbadf00d Jan 31 '21 at 18:27
  • (a) The proof of the algebraic version of the other direction (i.e. if "$A$ is surjective, then $A'$ is injective") I've mistakenly provided in my former response to (b) should hold for general normed vector spaces $X,Y$ and $T$. Do you agree? – 0xbadf00d Jan 31 '21 at 18:30
  • the implication ``$T'$ injective $\Rightarrow$ $T$ surjective'' is false as soon as $R(T)$ is not closed. – daw Feb 01 '21 at 07:47
  • In your answer you wrote that $\mathcal R(T)$ is closed by the closed graph theorem. Why is that true? Doesn't the closed graph theorem only imply that ${(x,Tx):x\in X}$ is closed? – 0xbadf00d Feb 01 '21 at 07:57
  • I meant the closed range theorem, sorry. – daw Feb 01 '21 at 08:20
  • Ah, I see.

    (Please note that there is still one line where you are writing closed graph theorem.)

     – 0xbadf00d Feb 01 '21 at 15:33
  • You wrote that injectivity is clear, but what is your argument? Is this correct: If $x\in\mathcal N(A)$, then $0=\langle Tx,y'\rangle=\langle x,T'y'\rangle$ for all $y'\in Y'$ and hence $x\in\mathcal R(T')\perp$. Now, since $T'$ is surjective, $\mathcal R(T')=X$ and hence $\mathcal R(T')\perp={0}$. Thus, $x=0$. – 0xbadf00d Feb 01 '21 at 15:35