TITLE: Exists a uniformly convex norm on Banach space satisfying certain condition?
QUESTION [8 upvotes]: Let $E$ be a Banach space with norm $\|\cdot\|$. Assume that there exists on $E$ an equivalent norm, denoted by $|\cdot|$, that is uniformly convex. Given any $k > 1$, does there exist a uniformly convex norm $\|\cdot\|'$ on $E$ such that$$\|x\| \le \|x\|' \le k\|x\| \text{ for all }x \in E?$$Progress. Maybe we could set$$\|x\|'^2 = \|x\|^2 + \alpha|x|^2$$with $\alpha > 0$ small enough? I am not sure what to do from there on out though.

REPLY [3 votes]: Lemma (Exercise 3.29, Brezis). Let $E$ be a normed vector space with a uniformly convex norm and fix $p > 1$. If $x$, $y \in \overline{N(0, M)} =: B$ are at least $\epsilon > 0$ apart, then there is some $\delta$ such that$$\left\|{{x + y}\over2}\right\|^p \le {{\|x\|^p + \|y\|^p}\over2} - \delta.$$

Suppose not for the sake of contradiction. Then there are sequences $(x_n)$ and $(y_n)$ in $B$ such that$$\|x_ - y_n\| > {{\|x\|^p + \|y\|^p}\over2} - {1\over n},$$and we may suppose that $|x_n| \to a$ and $y_n| \to b$ by compactness of $[0, M]$. Since$$a + b > \epsilon_0 \text{ and }{{a^p + b^p}\over2} \le \left({{a + b}\over2}\right)^p,$$we have that $a = b$. Let$$x_n' = {{x_n}\over{\|x_n\|}} \text{ and }y_n' = {{y_n}\over{\|y_n\|}}$$for which$${\epsilon\over a} + \text{o}(1) < \|x_n' - y_n'\| < 1 - \delta.$$Therefore,$$a^p + \text{o}(1) \le \left\|{{x_n + y_n}\over2}\right\|^p \le a^p(1 - \delta)^p + \text{o}(1),$$which is a contradiction.

Original Problem (Exercise 3.30, Brezis). If $E$ is a normed vector space that admits an equivalent uniform norm $|\cdot|_1$, then there is a uniformly convex norm $|\cdot|_2$ such that$$\|x\| \le |x|_2 \le k\|x\|$$for $k$ arbitrarily close to $1$.

Let$$|x|_2^2 = \|x\|^2 = \alpha|x_1|^2,$$then by the lemma,$$\|x\|^2 \le |x|_2 \le (1 + \alpha C^2)\|x\|^2,$$which can be made arbitrarily small and for any $x$, $y \in \overline{B(0, 1)}$ with respect to the new norm, then for any $|x - y|_2 > \epsilon$,$$\left|{{x + y}\over2}\right|_2^2 \le \left\|{{x + y}\over2}\right\|^2 + \left|{{x + y}\over2}\right|^2 \le \left\|{{x + y}\over2}\right\|^2 + {{|x|^2 + |y|^2}\over2} - \delta.$$