Convergence of Lipschitz Constants

Question

Let $V$ be a normed vector space and $F:V\rightarrow V$ Lipschitz, i.e. there exists $M\in[0,\infty)$ with $\|F(v_{1}) - F(v_{2})\| \leq M\,\|v_{1} - v_{2}\|$ for all $v_{1},v_{2}\in V$ (we let $M$ be the smallest constant with this property).

Moreover, let $\{F_{n}\}_{n=1,\ldots,\infty}$ be a sequence of functions $F_{n}:V\rightarrow V$ with

• there exists $M_{n}\in[0,\infty)$ with $\|F_{n}(v_{1}) - F_{n}(v_{2})\| \leq M_{n}\,\|v_{1} - v_{2}\|$ for all $v_{1},v_{2}\in V$ (again, we let $M_{n}$ be the smallest constant with this property);
• for all $v\in V$, we have $F_{n}(v)\rightarrow F(v)$ for $n\rightarrow \infty$ (point-wise convergence).

I am interested in showing $M_{n}\rightarrow M$ and here's my attempt: we have

$$\begin{aligned} \big|\; \| F_{n}(v_{1}) - F_{n}(v_{2}) \| - \| F(v_{1}) - F(v_{2}) \| \;\big| &\;\leq\; \big\| \big(F_{n}(v_{1}) - F_{n}(v_{2})\big) - \big(F(v_{1}) - F(v_{2})\big) \big\| \\ &\;\leq\; \big\| \big(F_{n}(v_{1}) - F(v_{1})\big) \big\| + \big\| \big(F_{n}(v_{2}) - F(v_{2})\big) \big\| \end{aligned}$$

where the last term converges to zero for all $v_{1},v_{2}\in V$. In other words

$$\| F_{n}(v_{1}) - F_{n}(v_{2}) \| - \| F(v_{1}) - F(v_{2}) \| \;\rightarrow\; 0$$

for all $v_{1},v_{2}\in V$, such that $M_{n}\rightarrow M$.

Do you agree with this line of reasoning? Is there an even shorter way to show convergence of the Lipschitz constants?

I feel there might be a mistake in using local convergence to obtain a result for the global property of Lipschitz continuity.

Answer 1

I don't believe it's true.

Let $c_0$ be the space of all sequences converging to $0$ equipped with the norm $\|\cdot\|_\infty$. Define $T_n : c_0 \to c_0$ as:

$$T_n(x_k)_{k=1}^\infty = x_ne_n = (\underbrace{0, \ldots, 0}_{n-1}, x_n, 0, \ldots)$$

for every $(x_k)_{k=1}^\infty \in c_0$.

$T_n$ is a bounded linear map so in particular it is a Lipschitz function, with the minimal Lipschitz constant $M_n$ equal to $\|T_n\| = 1$.

$(T_n)_{n=1}^\infty$ converges pointwise to the zero function $0 : c_0 \to c_0$. Indeed, for every $(x_k)_{k=1}^\infty \in c_0$ we have

$$T_n(x_k)_{k=1}^\infty = x_ne_n \xrightarrow{n \to \infty} 0$$

because $x_n \xrightarrow{n\to\infty} 0$.

However, $\|T_n\| =1$ for all $n \in \mathbb{N}$ so it does not converge to $0 = \|0\|$.

Answer 2

When you say "For a fixed $v\in V$..." above you mean for every fixed $v$, right?

It's not true that $M_n\to M$. For example, let $V=C([0,1])$, with $||f||=\sup_{t\in[0,1]}|f(t)|$, and define $F_n:V\to V$ by $$F_n(f)(t)=f(1/n)-f(0)\quad(t\in[0,1]).$$

Then $M_n=2$ for every $n$, but $F_n(f)\to0$ for every $f$, so $M=0$.

It's clear that $M\le\liminf_{n\to\infty}M_n$, but that's more or less all you can say.