Convergent Sequence in Normed Vector Space is Weakly Convergent

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Theorem

Let $\struct {X, \norm \cdot}$ be a normed vector space.

Let $x \in X$.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging to $x$.


Then $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.


Proof 1

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.

Let $f \in X^\ast$.

If $\norm f_{X^\ast} = 0$, then $f = 0$ and:

$\map f {x_n} \to \map f x$

Take $f \ne 0$, so that $\norm f_{X^\ast} \ne 0$.

Let $\epsilon > 0$.

We then have:

\(\ds \cmod {\map f {x_n} - \map f x}\) \(=\) \(\ds \cmod {\map f {x_n - x} }\) Definition of Linear Functional
\(\ds \) \(\le\) \(\ds \norm f_{X^\ast} \norm {x_n - x}\) Fundamental Property of Norm on Bounded Linear Functional

Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$:

there exists $N \in \N$ such that $\ds \norm {x_n - x} < \frac \epsilon {\norm f_{X^\ast} }$ for all $n \ge N$.

Then, for $n \ge N$, we have:

\(\ds \cmod {\map f {x_n} - \map f x}\) \(\le\) \(\ds \norm f_{X^\ast} \norm {x_n - x}\)
\(\ds \) \(<\) \(\ds \norm f_{X^\ast} \paren {\frac \epsilon {\norm f_{X^\ast} } }\)
\(\ds \) \(=\) \(\ds \epsilon\)

Since $\epsilon > 0$ was arbitrary, we obtain:

$\map f {x_n} \to \map f x$

Since $f \in X^\ast$ was arbitrary, we have:

$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.

$\blacksquare$


Proof 2

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.

Then, for each $f \in X^\ast$:

\(\ds \size {\map f {x_n} - \map f x}\) \(\le\) \(\ds \norm f_{X^\ast} \norm {x_n - x}_X\) Fundamental Property of Norm on Bounded Linear Functional
\(\ds \) \(\) \(\ds \stackrel{n \to \infty}{\longrightarrow} 0\)

$\blacksquare$


Sources

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