Let $(X,\left\|\cdot\right\|)$ be a separable normed space and $\left\{x_{n}\right\}$ be a dense subset of $\{x \in X:\|x\|=1\}$. Show that there is a sequence $\left\{f_{n}\right\}$ in $BL(X, \mathbb{K})$ such that $\left\|f_{n}\right\|=1$ for all $n$ and for every $x \neq 0$ in $X, f_{n}(x) \neq 0$ for some $n$. Moreover,
$$
\|x\|_{0}=\left(\sum_{n=1}^{\infty} \frac{\left|f_{n}(x)\right|^{2}}{2^{n}}\right)^{1 / 2}
$$
$\|\cdot\|_{0}$ defined above for $x \in X$ is a norm on $X$, in which $X$ is strictly convex and $\|x\|_{0} \leq\|x\|$, for all $x \in X$. Consequently, show there is an equivalent norm on $X$ in which $X$ is strictly convex.
I will be be using Hahn-Banach Extension theorem and the following lemma: Let X be a normed linear space over $\mathbb{K} = \mathbb{C}/\mathbb{R}$ and $0\ne a \in X$ , then there exists $f \in X' = BL(X,\mathbb{K})$ such that $f(a)=\|a\| \ne 0$ and $\|f\|_{op} = 1$.
My attempt: Construct a sequence of functions $f_n$ from the above lemma such that $f_n(x_n) = \|x_n\| = 1\ne 0$ for given sequence $x_n$ and $\| f_n\|_{op}=1$. Now, for every $x \ne 0 \in X$, there exists $m \in \mathbb{N}$ such that $x/\|x\|$ lies in a $\epsilon$-ball centered at $x_m$ where the linear functional $f_m$ is non zero as $f_m(x_m) \ne 0$ (I am a little unsure about this). This finishes the proof of the first part. Now we will show that $\| \cdot\|_{0}$ is a norm. Notice that $$ \|x\|_{0}=\left(\sum_{n=1}^{\infty} \frac{\left|f_{n}(x)\right|^{2}}{2^{n}}\right)^{1 / 2} \le \|x\|\left(\sum_{n=1}^{\infty} \frac{1}{2^{n}}\right)^{1 / 2} = \|x\|$$ If $x \ne 0$, then there is a $m \in \mathbb{N}$ such that $f_m(x) \ne 0$, which implies $\|x\|_{0} \ne 0$. If $x=0$, then $f_{n}(x)=0$ for all $n$ as they are linear, therefore $\|x\|_{0}=0$. Clearly $\|\lambda x\|_{0} = \lambda \|x\|_{0}$ (follows from linearity of $f_n$'s). I believe that the triangle inequality will follow from Minkowski's inequality (we also take the limit on dimension). If $\| x\|_{0}=\|y\|_{0}=1$ then we need to show $\|x_0+y_0\| <2$. I am stuck with this for sometime and the consequence part. Any hints? Are there any mistakes in my approach ?
Edit 1: I was able to show that $\| \cdot\|_0$ is strictly convex thanks to @daw. Now I am only stuck in showing that there is an equivalent norm in which $X$ is strictly convex. Consider the norm $\|\cdot\|_1$ defined by $\|x\|_1= \|x\|_0+\|x\|$ where $\|\cdot\|$ is the original norm on $X$. Then $\|x\| \le \|x\|_1 \le 2\|x\|$. Therefore, it is equivalent with $\|\cdot\|$. How do I show from here that $\|x\|_1$ is strictly convex?
Best Answer
A norm $\|\cdot\|$ is strictly convex if for all $x,y \in X$, with $x \ne y$, we have $\| \lambda x +(1-\lambda)y\| $ $<$ $ \lambda\|x\|+(1-\lambda)\|y\|$ for all $\lambda \in (0,1)$. Now, to show $\|\cdot\|_1$ is strictly convex, observe that $\| \lambda x +(1-\lambda)y\|_1=\| \lambda x +(1-\lambda)y\| +\| \lambda x +(1-\lambda)y\|_0$ is strictly less than $\lambda \|x\|+(1-\lambda)\|y\|+\lambda \|x\|_0+(1-\lambda)\|y\|_0 = \lambda \|x\|_1 + (1-\lambda)\|y\|_1$.