3
$\begingroup$

Let $X$ and $Y$ be two normed linear spaces. Let $T:X \to Y^*$ be a linear operator (not necessarily continuous) and let $T^*$ be its adjoint, i.e. $T^*:Y \to X^*$ is defined by

$ \langle T^*y,x \rangle := \langle Tx,y \rangle ,$

where $\langle \cdot , \cdot \rangle$ is a proper duality pairing (i.e. between $X^*$ and $X$ on the left hand side and between $Y^*$ and $Y$ on the right hand side). Let $R(T)$ denote the range of $T$. Is it true that:

$T$ is injective on $R(T)$ if and only if $T^*$ is surjective onto $X^*$?

$\endgroup$
3
$\begingroup$

The full equivalence does not hold in general. Rather it holds: $T$ is injective if and only if $R(T^*)$ is weak-star dense in $X^*$.

If $T^*$ is surjective, then $T$ is injective: Suppose $Tx=0$, then $$ 0 = \langle Tx,y\rangle_{Y^*,Y} = \langle T^*y,x\rangle_{X^*,X}, $$ and since $R(T^*)=X^*$, it follows $0= \langle x^*,x\rangle_{X^*,X}$ for all $x^*\in X^*$. Hence $x=0$.

Here, it would have been sufficient to suppose that $R(T^*)$ is dense in $X^*$.

Now, let $T$ be injective, assume that $R(T^*)$ is not weak-star dense in $X^*$. Then there is $x^*\in X^* \setminus \overline{R(T^*)}^{w^*}$, where closure is taken with respect to the weak-star topology. By Hahn-Banach separation theorem, there is $x\in X$, $t\in\mathbb R$, such that $$ \langle x^*,x\rangle_{X^*,X} < t \le \langle T^*y,x\rangle_{X^*,X} \quad \forall y\in Y. $$ Since $T^*$ is linear, it follows $\langle T^*y,x\rangle_{X^*,X}=0$ for all $y\in Y$, which is equivalent to $\langle Tx,y\rangle_{Y^*,Y}=0$ for all $y$. Hence $Tx=0$ and by injectivity $x=0$ follows, which is a contradiction.

As a counter-example, which shows that injectivity of $T$ does not implies surjectivity of $T^*$, you can choose any compact and injective $T$, with $X,Y$ being infinite-dimensional. Then $T^*$ is compact as well, but $R(T^*)$ cannot be closed.

$\endgroup$
  • $\begingroup$ Two comments for this application of Hahn-Banach separation theorem: 1. The theorem states that there exist a $x^{**} \in X^{**}$ s.t. ... Hence you have additionally assumed the reflexivity of $X$. 2. The theorem states that there exists a $t\in \mathbb{R}$ s.t. the separation property holds, not necessarily $t=0$. (see H-B theorem for instance here en.wikipedia.org/wiki/…) $\endgroup$ – Wojtek Feb 9 '15 at 11:11
  • $\begingroup$ 1) H-B works in topological vector spaces as well, so we can really choose $x\in X$. 2) see edit: inserted $t$. $\endgroup$ – daw Feb 9 '15 at 12:31
  • $\begingroup$ Thanks for answering my comment. $\endgroup$ – Wojtek Feb 10 '15 at 9:37
  • $\begingroup$ Thanks for answering my comment. 1) Unless there is a version of H-B dealing with predual spaces I am not aware of, I don't agree with your explanation. The theorem, considering $X^*$ as a topological vector space, only gives you an element $x^{**}\in X^{**}$ such that $\langle x^{**}, x^* \rangle < t \leq \langle x^{**}, T^* y \rangle $. Representing this $x^{**}$ as some $x\in X$, as stated above, is nothing else but saying that $X$ is reflexive. 2) If we have $t$ instead of $0$ then it is not true that $\langle T^*y,x \rangle =0$ for all $y\in Y$ (take, for instance, $t>0$). $\endgroup$ – Wojtek Feb 10 '15 at 9:45
  • $\begingroup$ 1) Talking about topological vector spaces, the dual of $X^*$ ($X^*$ equipped with weak-star topology) is $X$. 2) $y$ is arbitrary from the vector space $Y$. If $\langle T^*y,x\rangle$ would be non-zero for some $y$, then you can take $sy$, $s\in \mathbb R$ to obtain a contradiction. $\endgroup$ – daw Feb 10 '15 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.