Convergent Sequence in Normed Vector Space is Weakly Convergent/Proof 1
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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
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 $x$.
$\blacksquare$