Condition for which $T'$ is injective. (Help understanding an incredibly terse proof.) Working from Hilbert Space Methods for Partial Differential Equations by Showalter. The proof of one of the theorems refers to a prior result that seems to me to be almost unrelated to the theorem which is to be proved:

Theorem I.5.A (page 17). If $V$ is a linear space, $W,q$ is a seminorm space and $T \in L(V,W)$ has dense range, then $T'$ is injective on $W'$.
Proof. [The result] follows from Section I.3.2.

Section I.3.2 (pages 10-11) lists a theorem and a lemma:

Lemma. Let $T: D \to W$ be given, where $D$ is a subset of a seminormed space $V,p$ and $W,q$ is a normed linear space. There is at most one continuous $\overline{T}: \overline{D} \to W$ for which $\overline{T}|_D = T$.


Theorem I.3.A Let $T \in \mathscr{L}(D,W)$ where $D$ is a subspace of the seminormed space $V,p$ and $W,q$ is a Banach space. Then there exists a unique $\overline{T} \in \mathscr{L}(\overline{D},W)$ such that $\overline{T}|_D = T$ and $|\overline{T}|_{p,q} = |T|_{p,q}$.

I just don't see at all how Theorem I.5.A follows from either the Lemma or from Theorem I.3.A!
My guess is the set $D$ in Theorem I.3.A translates into Range$(T)$ in Theorem I.3.A. And The uniquess result for $\overline{T}$ in Theorem I.3.A somehow translates into the injectivity of $T'$ in Theorem I.5.A. But maybe that's totally the wrong idea? Or if it is the right idea, I have no idea what the connection is?
Note: in this notation, $L(V,W)$ refers to all linear operators and $\mathscr{L}(V,W)$ refers to bounded linear operators. $|T|_{p,q}$ refers to the operator seminorm induced by seminorms $p$ and $q$.
 A: This does not require the Lemma or Theorem 1.3.A. Yo only need the deifinition of $T'$.
If $T'w'=0$ then $T'w'(x)=0$ for all $x \in V$. By definition this means $w'(Tx)=0$ for all $x\in V$. By continuity of $w'$ and the fact that $\{Tx: x\in V\}$ is dense in $W$ we get $w'(w)=0$ for all $w \in W$, so $w'=0$. This proves that $T'$ is injective.
A: Original asker here. Thank you to geetha290krm for a concise proof. I also came up with a proof by explicitly chasing definitions around, which I figured I'd write up and contribute as well:
(1) Let us rephrase the statement to be proved. The given hypotheses are:

*

*$V$ is a linear space.

*$W,q$ is a seminorm space.

*$T \in L(V,W)$ has dense range.

The claim to be proved is that $T'|_{W'}$ is injective. Thus, we need to show that if $T'f_1 = T'f_2$ for $f_1, f_2 \in \mathscr{L}(W,\mathbb{R})$, then $f_1 = f_2$. So, to the above hypotheses, we add the hypothesis that

*

*$T'|_{W'}f_1 = T'|_{W'}f_2$ and $f_1, f_2 \in \mathscr{L}(W,\mathbb{R})$,

and we want to show that $f_1 = f_2$.
(2) Let us carefully unpack the definitions in the added hypothesis:

*

*Because $T'f_1$ and $T'f_2$ are linear functionals on
$V$, to say that $T'f_1 = T'f_2$ is to say that
$(T'f_1)(v) = (T'f_2)(v)$ for all $v \in V$.


*By definition of the transpose map $T'$, this means that $(f_1 \circ
   T)(v) = (f_2 \circ T)(v)$ for all $v \in V$, which means that $f_1(Tv) = f_2(Tv)$ for all $v \in V$.
So, the hypothesis $T'|_{W'}f_1 = T'|_{W'}f_2$ implies that $f_1$ and $f_2$ agree on the range of $T$, which we denote $R(T)$. In short, $f_1|_{R(T)} = f_2|_{R(T)}$.
(3) By hypothesis, $R(T)$ is dense in $W$. We'd like to invoke Theorem I.3.A. To do this, let's rewrite Theorem I.3.A using terminology appropriate to the current claim: Let $f|_{R(T)} \in \mathscr{L}(R(T), \mathbb{R})$ where $R(T)$ is a dense subset of seminormed space $W$ and $\mathbb{R}$ is (obviously) a Banach space. Then there exists a unique $f \in \mathscr{L}(W, \mathbb{R})$ such that $f(w) = f|_{R(T)}(w)$ for all $w \in R(T)$.
(4) By (3), the functional $f_1|_{R(T)}: R(T) \to \mathbb{R}$ can be uniquely extended to $f_1: W \to \mathbb{R}$, and $f_2|_{R(T)}$ can be similarly extended to $f_2$. But since the extensions $f_1$ and $f_2$ are unique, we have that $f_1(w) = f_2(w)$ for all $w \in W$, i.e. that $f_1 = f_2$.
To conclude, we have shown that for the given hypotheses, if $T'|_{W'}f_1 = T'|_{W'}$ then $f_1 = f_2$, meaning that $T'$ is injective on $W'$.
