I was reading about uniformly convex Banach spaces from the book "Topics in metric fixed point theory" by Goebel and Kirk. As a reference, I just mention the definition of a uniformly convex Banach space, which I took from the original paper of Clarkson, where it was introduced. It turns out that this (original) definition and that given in the book by Goebel are equivalent. However, I will be using the one used by Clarkson.
Definition (Uniform Convexity): A Banach space $\left( X, \| \cdot \| \right)$ is said to be uniformly convex if for given $0 < \varepsilon \leq 2$, there is some $\delta > 0$ for all $x, y \in X$ with $\| x \| \leq 1$, $\| y \| \leq 1$ and $\| x – y \| \geq \varepsilon$, we have $\| \frac{x + y}{2} \| \leq 1 – \delta$.
Based on this definition, we then get the definition of modulus of convexity, which is defined simply as a function $\delta: \left[ 0, 2 \right] \rightarrow \left[ 0, 1 \right]$, as
$$\delta \left( \varepsilon \right) = \inf \left\lbrace 1 – \left| \left| \dfrac{x + y}{2} \right| \right| : \| x \| \leq 1, \| y \| \leq 1 \text{ and } \| x – y \| \geq \varepsilon \right\rbrace.$$
Now, in the book it is given that the following proposition is an easy exercise for the reader.
Proposition: If $\left( x_n \right)$ and $\left( y_n \right)$ are sequences in a uniformly convex Banach space $\left( X, \| \cdot \| \right)$ such that $\lim\limits_{n \rightarrow \infty} \| x_n \| = \lim\limits_{n \rightarrow \infty} \| y_n \| = \lim\limits_{n \rightarrow \infty} \dfrac{1}{2} \| x_n + y_n \| = 1$, then $\lim\limits_{n \rightarrow \infty} \| x_n – y_n \| = 0$.
However, I am not able to prove this proposition. There are two major difficulties I am facing to start the proof:-
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$\lim\limits_{n \rightarrow \infty} \| x_n \| = 1$ does not imply that after a stage, he sequence of norms $\| x_n \|$ would satisfy $\| x_n \| \leq 1$. Therefore, we cannot use the definition of uniform convexity and the method of contradiction to prove the result.
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How do we know that the limit $\lim\limits_{n \rightarrow \infty} \| x_n – y_n \|$ exists?
Any help will be appreciable!
Best Answer
Let $u_n=\frac {x_n} {\|x_n\|}$ and $v_n=\frac {y_n} {\|y_n\|}$. Let us show that $\|\frac {u_n+v_n} 2\| \to 1$. For this note that $\|x_n-u_n\|=\|x_n-\frac {x_n} {\|x_n\|} \|=|1-\|x_n\|| \to 0$. Similarly, $\|y_n-v_n\|\to 0$, Hence $|\|\frac {u_n+v_n} 2\| -\|\frac {x_n+y_n} 2\| |\leq \frac {\|x_n-u_n\|+\|y_n-v_n\|} 2 \to 0$. This proves that $\|\frac {u_n+v_n} 2\| \to 1$.
Now suppose $\|u_n-v_n\| \geq \epsilon$ for infinitely many $n$. Then $\delta(\epsilon) \leq 1-\|\frac {u_n+v_n} 2\| \to 0$ (through a subsequence), a contradiction. So we have proved that $\|u_n-v_n\| <\epsilon$ for $n$ sufficiently large . Since $\|x_n-u_n\| \to 0$ and $\|y_n-v_n\| \to 0$ we get $\|x_n-y_n\| \to 0$.