Weakly Convergent Sequence in Normed Vector Space is Bounded
Theorem
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weakly convergent sequence in $X$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is bounded.
Proof
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $X$.
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $X$.
Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation on $X$.
Let:
- $x^\wedge = \map J x$
for each $x \in X$.
Consider the sequence $\sequence {x_n^\wedge}_{n \mathop \in \N}$ in $X^{\ast \ast}$.
Then, for each $f \in X^\ast$, we have:
- $\map {x_n^\wedge} f = \map f {x_n}$
Since $\sequence {x_n}_{n \mathop \in \N}$ converges weakly we have that:
- $\sequence {\map f {x_n} }_{n \mathop \in \N}$ converges for each $f \in X^\ast$.
So, from Convergent Complex Sequence is Bounded, $\sequence {\map f {x_n} }_{n \mathop \in \N}$ is bounded for each $f \in X^\ast$.
So for each $f \in X^\ast$ there exists a real number $M_f > 0$ such that:
- $\cmod {\map f {x_n} } \le M_f$
for each $n \in \N$.
That is:
- $\cmod {\map {x_n^\wedge} f} \le M_f < \infty$
so:
- $\ds \sup_{n \mathop \in \N} \cmod {\map {x_n^\wedge} f}$ is finite for each $f \in X^\ast$.
From Normed Dual Space is Banach Space, $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ is a Banach space, and so we may apply the Banach-Steinhaus Theorem.
From the Banach-Steinhaus Theorem, we then obtain:
- $\ds \sup_{n \mathop \in \N} \norm {x_n^\wedge}_{X^{\ast \ast} }$ is finite.
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have:
- $\norm {x_n^\wedge}_{X^{\ast \ast} } = \norm {x_n}_X$
So:
- $\ds \sup_{n \mathop \in \N} \norm {x_n}_X$ is finite.
So we may conclude that:
- $\sequence {x_n}_{n \mathop \in \N}$ is bounded.
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $27.1$: Weak Convergence