Limit Inferior of Norm of Weakly Convergent Sequence is Bounded Below by Norm of Weak Limit

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:


 * $\ds \norm x \le \liminf_{n \mathop \to \infty} \norm {x_n}$