Limit Inferior of Norm of Weakly Convergent Sequence is Bounded Below by Norm of Weak Limit
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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 weakly to $x$.
Then, we have:
- $\ds \norm x \le \liminf_{n \mathop \to \infty} \norm {x_n}$
Proof
From Existence of Support Functional, there exists $f \in X^\ast$ such that:
- $\map f x = \norm x$
and:
- $\norm f_{X^\ast} = 1$
Since:
- $x_n \weakconv x$
we have:
- $\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f x$
That is:
- $\ds \lim_{n \mathop \to \infty} \map f {x_n} = \norm x$
From Absolute Value of Limit, we have:
- $\ds \lim_{n \mathop \to \infty} \cmod {\map f {x_n} } = \norm x$
From Convergence of Limsup and Liminf, we have:
- $\ds \liminf_{n \mathop \to \infty} \cmod {\map f {x_n} } = \norm x$
From Fundamental Property of Norm on Bounded Linear Functional, we have:
- $\cmod {\map f {x_n} } \le \norm f_{X^\ast} \norm {x_n}$
Since:
- $\norm f_{X^\ast} = 1$
we have:
- $\cmod {\map f {x_n} } \le \norm {x_n}$
So, from Inequality Rule for Real Sequences:
- $\ds \liminf_{n \mathop \to \infty} \cmod {\map f {x_n} } \le \liminf_{n \mathop \to \infty} \norm {x_n}$
That is:
- $\ds \norm x \le \liminf_{n \mathop \to \infty} \norm {x_n}$
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $27.1$: Weak Convergence