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 $X^*$?


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.

  • $\begingroup$ 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

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