Theorem 3.10 in Rudin's Functional Analysis

Question

Theorem 3.10: Suppose $X$ is a vector space and $X'$ is a separating vector space of linear functionals on $X$. Then the $X'$-topology $\tau'$ makes $X$ into a locally convex space whose dual space is $X'$.

In addition to having the same trouble as this questioner, I am also having trouble with this:

Conversely, suppose $\Lambda$ is a $\tau '$-continuous linear functional on $X$. Then $|\Lambda x| < 1$ for all $x$ in some set $V$ of the form (1). Condition (b) of Lemma 3.9 therefore holds; hence so does (a): $\Lambda = \sum \alpha_i \Lambda_i$.

(1) refers to $V = \{x : |\Lambda_i x| < r_i ~~\text{ for }~~ 1 \le i \le n \}$ and Lemma 3.9 says the following:

Lemma 3.9: Suppose $\Lambda_1,...,\Lambda_n$ and $\Lambda$ are linear functionals on a vector space $X$. Let $$N = \{x : \Lambda_1 x = ... = \Lambda_n x = 0\}.$$ The following three are then equivalent

(a) There are scalars $\alpha_1,...,\alpha_n$ such that $$\Lambda = \alpha_1 \Lambda_1 + ... + \alpha_n \Lambda_n$$

(b) There exists $\gamma < \infty$ such that $$|\Lambda x| \le \gamma \max_{1 \le i \le n} |\Lambda_i x| ~~~(x \in X)$$

(c) $\Lambda x = 0$ for every $x \in N$.

My question is, why is condition (b) satisfied?

Answer 1

I complete the proof that was suggested by logarithm in the comment.

I start from \begin{equation*} V = \{x\in X:|\Lambda_ix|<r_i\text{ for }1\le i\le n\} \subset \{x\in X: |\Lambda x| < 1\}. \end{equation*} Put $\gamma = 2/\min\{r_1,\dots,r_n\}$ and $M_x = \max\{|\Lambda_1x|,\dots,|\Lambda_nx|\}$ for $x\in X$. If $M_x = 0$ for all $x\in X$, then $V = X \subset \{x:|\Lambda x|<1\}$ implies that $|\Lambda x| = 0 = \gamma M_x$ for all $x\in X$.

Assume that $M_x > 0$ for some $x\in X$. If $M_x > 0$, then $|\Lambda_i(x/(\gamma M_x))| = |\Lambda_ix|/(\gamma M_x) \le 1/\gamma < r_i$ for all $i$; thus $x/(\gamma M_x) \in V$ and $|\Lambda x| = \gamma M_x|\Lambda (x/(\gamma M_x))| < \gamma M_x$.

If $M_x = 0$, then there is $y\in X$ with $M_y >0$ by the assumption. Let $0 = t_1 < t_2 < \cdots < 1$ with $t_j\to 1$ as $j\to\infty$ and $x_j = t_jx + (1-t_j)y$. Since \begin{align} M_{x_j} &\ge |\Lambda_ix_j| = |t_j\Lambda_ix + (1-t_j)\Lambda_iy| \ge (1-t_j)|\Lambda_iy| \\ M_{x-x_j} &\ge |\Lambda_i(x-x_j)| = |(1-t_j)\Lambda_i(x-y)| \ge (1-t_j)|\Lambda_iy| \end{align} for all $i$ and $j$, $M_{x_j}\ge (1-t_j)M_y>0$ and $M_{x-x_j}\ge (1-t_j)M_y>0$ for all $j$. Thus, \begin{align*} |\Lambda x| &\le |\Lambda(x-x_j)| + |\Lambda x_j| \le \gamma (M_{x-x_j} + M_{x_j}) \\ &\le \gamma [(1-t_j)(M_x + M_y) + t_j M_x + (1-t_j)M_y] \\ &= 2\gamma(1-t_j)M_y \to 0 \end{align*} as $j\to\infty$. Therefore, $|\Lambda x| = 0 = \gamma M_x$.

Answer 2

Here's an attempt to prove the equivalent condition (c).

More explicitly, I attempt to prove: $$ N=\bigcap_{i}\ker\Lambda_i \subset \ker \Lambda $$ Let $V$ be a basic neighborhood of $0$ of the sort: $$ V=\bigcap_{i} \{\ x:|\Lambda_i x|<r_i \ \} $$ Assume V satisfies: $$ |\Lambda x| \leq 1 \ \ \ :\ \ \ x\in V $$

Obviously, $N\subset V$

Suppose there exist a $x\in N$ such that: $$|\Lambda x|=c>0$$ Since $N$ is a subspace, for each $n\in \mathbb{N}$, we have: $$ nx\in N \subset V $$ Therefore: $$ |\Lambda nx| = nc \underset{n\to \infty}{\to} \infty $$ This is absurd, since $\Lambda$ is bounded in $V$. This completes the proof of condition (c), which implies (b) and (a).