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Denote $$L(f) = \sup_{p,q \in X}{\dfrac{\rho(f(p),f(q))}{\rho(p,q)}}$$

The following proposition is extracted from the book 'Lipschitz Algebras' by Weaver.

Proposition $1.2.4$ Let $X$ and $Y$ be metric spaces and let $f$ and $f_i (i \in I)$ be functions from $X$ to $Y$. Suppose $f_i \rightarrow f$ pointwise. Then $L(f) \leq \sup L(f_i)$.

Proof: For any $p,q \in X$, we have

$$\rho (f(p),f(q)) = \lim \rho(f_i(p),f_i(q)),$$

and dividing by $\rho(p,q)$ shows that $\rho(f(p),f(q))/\rho(p,q) \leq \sup L(f_i)$. Taking the supremum over $p$ and $q$ now yields the desired conclusion.

Question: How to obtain the inequality $\rho(f(p),f(q))/\rho(p,q) \leq \sup L(f_i)$?

Idonknow
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1 Answers1

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Let, $M=\sup_{i} L(f_i);$ Note that by definition of $L(f)$, for each $i$, we have

$\rho (f_i(p),f_i(q)) \leq L(f_i) \rho(p,q) \leq \big(\sup_{i} L(f_i)\big)\rho(p,q) = M .\rho(p,q)$

Now Left side of the equation converge to $\rho(f(p),f(q))$ as $i\to \infty$. Hence we have

$\rho (f(p),f(q)) \leq M. \rho(p,q)$ and this holds for every $p,q \in X.$ So we have $\sup _{p,q \in X} \frac{\rho(f(p),f(q))}{\rho (p,q)} \leq M$ which gives us $L(f) \leq M$.

Timon
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