# Is the injectivity of the operator equivalent to the surjectivity of its adjoint

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.

• 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/…) – Wojtek Feb 9 '15 at 11:11
• 1) H-B works in topological vector spaces as well, so we can really choose $x\in X$. 2) see edit: inserted $t$. – daw Feb 9 '15 at 12:31
• Thanks for answering my comment. – Wojtek Feb 10 '15 at 9:37
• 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, $t>0$). – Wojtek Feb 10 '15 at 9:45
• 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. – daw Feb 10 '15 at 11:54