I want to prove the following:
Let X be a uniformly convex Banach space, $\{x_n\}_{n\in\mathbb{N}} \subset X, x \in X, \|x_n\| \to \|x\|$, $x_n$ converges weakly to $x$. Then $x_n$ also converges strongly to $x$.
Actually I have not really an idea how to do this.
Would be grateful for some tips!
S
I mean what I have is, that $ < f, x_n -x> \to 0$ $\forall f \in X^*$ and $\|x_n\| \to \|x\|$.
And $\forall \epsilon > 0 \exists \delta > 0 \forall x,y \in X, \|x\|,\|y\| \leq 1: \|x-y\|\geq\epsilon \Rightarrow \frac{\|x+y\|}{2} \leq 1-\delta $
or $\forall \epsilon > 0 \exists \delta > 0 \forall x,y \in X, \|x\|,\|y\| \leq 1:\frac{\|x+y\|}{2} > 1-\delta \Rightarrow \|x-y\| < \epsilon $.
And I want to show that $\|x_n-x\|\to 0$.
My first idea was that I want to have somehow $\|x_n-x\|<\epsilon$. So I need to fulfill the condition of the uniformly convexness?
EDIT - Possible Solution:
Let $\epsilon > 0$ arbitrary. As $X$ is a uniformly convex Banach space there is a $\delta > 0$ such that for every $x,y \in X$ with $\|x\|,\|y\|\leq 1: \frac{\|x+y\|}{2} > 1-\delta$ implies $\|x-y\|<\epsilon$.
Let $f\in X^*$ such that $f(x)=\langle f, x\rangle = \|x\|$ and $\|f\|_* = 1$.
Then we have $\frac{\|x_n+x\|}{2} = \frac{\|f\|_{*}\|x_n+x\|}{2} \geq \langle f,\frac{x_n+x}{2}\rangle \to \langle f,x\rangle = \|x\| = 1 > 1-\delta$.
So $\|x_n-x\| < \epsilon$.
As $\epsilon$ was arbitrary $\{x_n\}_{n\in\mathbb{N}}$ converges strongly to x.
But then I didn't use that $\|x_n\| \to \|x\|$.
Assume that $X$ is uniformly convex, $x_n \overset{\text{weakly}}{\longrightarrow} x$ and $\| x_n \| \longrightarrow \| x\|$.
Now let $y_n = \frac{x_n}{\|x_n\|}$ and $y = \frac{x}{\|x\|}$. Then we have $\| y_n \| = 1$ and $\| y \| = 1$ and also $y_n \overset{\text{weakly}}{\longrightarrow} y$ and $\| y_n \| \longrightarrow 1$. Now, by Hahn-Banach, take $f \in X'$ with $\| f \| = 1$ and $|f(y)| = 1$. Then we get, due to weak convergence: $|f(y_n)| \longrightarrow 1$. Therefore, we obtain $\left|\frac{f(y_n) + f(y)}{2}\right| \longrightarrow 1$ and also $\left|\frac{f(y_n) + f(y)}{2}\right| \leq \| f \| \left\| \frac{y_n + y}{2} \right\| = \left\| \frac{y_n + y}{2} \right\| \leq 1$. By the sandwich principle, we have that $\left\| \frac{y_n + y}{2} \right\| \longrightarrow 1$. Now, uniform convexity of $X$ implies that $\| y_n - y \| \longrightarrow 0$, meaning $y_n \longrightarrow y$ and due to $\| x_n \| \longrightarrow \| x \| $, we obtain that $x_n \longrightarrow x$.