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.