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Let $c_0$ be a spaces of sequences converging to $0$ with the following norm $$ \|x\|=\sup\{|x_i|: i\in \mathbb{N}\}+\left(\sum_{i=1}^{\infty}\left(\frac{x_i}{i}\right)^2\right)^{\frac{1}{2}} $$ Prove that $(c_0,\|.\|)$ is strictly convex but not uniformly convex.

Thank you for your kind help.

I can prove that $(c_0,\|.\|)$ is not uniformly convex by choosing $\{x^k\}, \{y^k\}\subset c_0$ given by $$ x^k_i= \begin{cases} \frac{1}{2}\sqrt{1-\frac{1}{(k+1)^2}-\frac{1}{(k+2)^2}}& \text{if}\quad i=1\\ \frac{1}{2}& \text{if}\quad i\in\{k+1, k+2\} \\ 0&\text{if}\quad\text{otherwise} \end{cases} $$ $$ y^k_i= \begin{cases} \frac{1}{2}\sqrt{1-\frac{1}{k^2}-\frac{1}{(k+2)^2}}& \text{if}\quad i=1\\ \frac{1}{2}& \text{if}\quad i\in\{k, k+2\} \\ 0&\text{if}\quad\text{otherwise} \end{cases} $$ We are easy to check $\|x^k\|=\|y^k\|=1$, $\|x^k-y^k\|\geq \frac{1}{2}$ but $$ \left\|\frac{x^k+y^k}{2}\right\|\longrightarrow 1 \quad\text{as}\quad k\longrightarrow \infty. $$

impartialmale
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2 Answers2

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It may be helpful to see the general structure of what goes on here. Suppose $X$ is a normed space, $Y$ is a strictly convex normed space, and $T:X\to Y$ is an injective bounded operator. Then the norm $$\|x\|=\|x\|_X+\|Tx\|_Y \tag{1}$$ on $X$ is equivalent to the original norm $\|\cdot \|_X$, and is strictly convex. The equivalence is clear: $$\|x\|_X\le \|x\| \le (1+\|T\|)\|x\|_X$$ For strict convexity, suppose $\|x+y\|=\|x\|+\|y\|$. In view of (1) and the triangle inequality, this implies $\|Tx +Ty\|_Y =\|Tx\|_Y + \|Ty\|_Y $. Since $Y$ is strictly convex, the vectors $Tx$ and $Ty$ are collinear: formally, $\alpha Tx = \beta Ty$ for some nonzero $\alpha,\beta$. But $T$ is injective, hence $\alpha x = \beta y$, which was to be proved.

The above applies to $c_0$ because $(x_i)\mapsto (x_i/i)$ is bounded from $c_0$ to $\ell_2$, the latter being strictly convex.

However, there is no uniformly convex norm on $c_0$, because uniformly convex spaces are reflexive and $c_0$ is not: $c_0^{**} = \ell_1^* = \ell_\infty$.

user127096
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Following the guidance of Robert Israel on the link Strict convexity of a norm on $C[0,1]$ we can prove that $(c_0, \|.\|)$ is strictly convex. Indeed, suppose that $x, y\in c_0\setminus\{0\}$ such that $\|x+y\|=\|x\|+\|y\|$. Consider the inner product and the norm on $c_0$ given by $$ \langle x,y\rangle=\sum_{i=1}^{\infty}\frac{x_iy_i}{i^2}, $$ $$ \|x\|_1=\sqrt{\langle x,x\rangle}. $$ Since $\|x+y\|=\|x\|+\|y\|$, it follows that $\|x+y\|_1=\|x\|_1+\|y\|_1$ If $x\ne\lambda y$ for all $\lambda\in\mathbb{R}$ then $$ \langle x-\lambda y, x-\lambda y \rangle>0, $$ $$ \langle x+\lambda y, x+\lambda y \rangle>0 $$ for all $\lambda\in\mathbb{R}$. Expanding the following inequalities we obtain $|\langle x, y\rangle|<\|x\|\|y\|$ It follows that $\|x+y\|_1<\|x\|_1+\|y\|_1$, an absurd.

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impartialmale
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