Mazur Lemma for sequence of Lipschitz functions

Question

This question regards to the Proposition 5.1 in Godefroy and Kalton (2003). Let $E$ be a finite-dimension Banach space and let $(L_n)_{n\in \mathbb{N}}$ be a sequence of functions $L_n: E\to \mathcal{F}(E)$ with finite-dimensional range, where $\mathcal{F}(E)$ is the Lipschitz-free space of $E$, that is, the canonical predual of $\mathrm{Lip}_0(E)$, which is the Banach space of Lipschitz functions of the form $f:E\to \mathbb{R}$ that vanish at $0$, equiped with the norm $$\|f\|_\mathrm{Lip} = \sup \left\{\frac{|f(x)-f(y)|}{\|x-y\|}: x\neq y \in E\right\}.$$ Suppose that, given an $\epsilon > 0$, $(L_n)_{n\in\mathbb{N}}$ be such that $L_n(0) = 0$, $\lim_n \langle f, L_n(x)\rangle = f(x)$ for $x\in B_E$ and $\lim\sup_n \|L_n\|_{\mathrm{Lip}} < 1 + \epsilon$. Since $\lim_n \langle f, L_n(x)\rangle = f(x)$, we know that $L_n$ weakly converges to
the evaluation functional $\delta_x \in \mathcal{F}(E)$, for every $x\in B_E$. Fixing $x \in B_E$, by Mazur's lemma, there's a sequence of convex combinations of $L_n(x)$'s that strongly converges to $\delta_x$.

How can I guarantee that there is a sequence of Lipschitz functions $H_n$ with finite-dimensional range such that $H_n(0) = 0$, $H_n(x) \to \delta_x$ for $x\in B_E$ and $\|H_n\|_{\mathrm{Lip}} < 1+ 1/n$ using convex combinations?

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I am having trouble with the Lipschitz norm of $H_n$. I know that there is a sequence of $L_n$'s with that property, using a diagonal argument with sequences $(L_{n,m})$ such that $\lim\sup_n\|L_{n,m}\|<1+1/m$.

For my attempt, if I wanted to prove that there is sequence of convex combination of $L_n$'s that covers all $x\in B_E$, I would have to prove that $\delta(B_E) \subset \cap_{x\in B_E} \mathrm{conv} \{L_n(x)\}$. But that's too restricted. Defining $H_n$ by creating different sequences of convex combinations for every $x\in B_E$, I tried to use triangular inequalities and the fact that $L_n(0) = 0$, but I only got that the norm is smaller than something bigger than 2 or 4.

Answer 1

Using a diagonal argument, we can consider $(L_n)_{n\in\mathbb{N}}$ such that $\|L_n\|_{\mathrm{Lip}} < 1 + 1/n$ for every $n\in \mathbb{N}$.

The secret to guarantee the Lipschitz constant for the new sequence $H_n$ relies on another diagonal argument: fixed $x\in B_E$, then for every $k\in \mathbb{N}$, we can use the Mazur Lemma for the sequence $(L_n(x))_{n\geq k}$ to obtain a sequence of convex combinations of $L_n(x)$, $(H_{k,n}(x))_{n\in\mathbb{N}}$, strongly convergent to $x$. Then we consider the diagonal sequence $(H_{n}(x))_{n\in\mathbb{N}} := (H_{n,n}(x))_{n\in\mathbb{N}}$.

Repeating the process for every $x\in B_E$, and leaving $H_n(x) = L_n(x)$ when $x\in E\setminus B_E$ (since here we don't have weak convergence), give us a sequence $(H_{n})_{n\in\mathbb{N}}$ of functions with finite-dimensional range such that $H_n(0) = 0$ and $H_n(x)\to \delta_x$ for $x\in B_E$. Now for the Lipschitz constant, let $x,y\in E$ and $n\in\mathbb{N}$. Then we have \begin{equation} \|H_n(x)-H_n(y)\|_{\mathcal{F}} = \left\Vert \sum_{m\geq n}\lambda_m(x) L_m(x) - \sum_{m\geq n}\lambda_m(y) L_m(y)\right\Vert_{\mathcal{F}}, \end{equation} where $\sum_{m\geq n}\lambda_m(z) = 1$ and $(\lambda_m(z))_{m\geq n}$ is almost null for every $z\in B_E$. Now, \begin{align} \|H_n(x)-H_n(y)\|_{\mathcal{F}} &\leq \left\Vert \sum_{m\geq n}\lambda_m(x) [L_m(x) - L_m(y)]\right\Vert_{\mathcal{F}} + \left\Vert \sum_{m\geq n}[\lambda_m(x) -\lambda_m(y)] L_m(y)\right\Vert_{\mathcal{F}}\\ & < \left|\sum_{m\geq n} \lambda_m(x) (1+1/m)\right|\|x-y\| + \left|\sum_{m\geq n}[\lambda_m(x) -\lambda_m(y)]\right|\|L_n\|_{\mathrm{Lip}}\\ & < (1+1/n)\|x-y\|. \end{align} Thus $\|H_n\|_{\mathrm{Lip}} < 1 + 1/n$.