Let $K$ be a compact set of a normed space $(X, \|\cdot\|)$. Suppose $f \colon K \to \mathbb R$ satisfies
$$
\frac{|f(x) – f(y)|}{\|x-y\|} < 1
\qquad \text{for all } x, y \in K \text{ with $x \neq y$}.
$$
Then $f$ is Lipschitz continuous with Lipschitz constant $1$.
Q1: Is $f$ also Lipschitz continuous with some Lipschitz constant $L<1$?
Q2: Does the answer remain the same if we replace $X$ and $R$ by generic metric spaces?
Assuming $K$ to be compact is necessary: For $K = \mathbb R$ any $f \in C^1(\mathbb R)$ with $\|f'\|_{L^\infty(\mathbb R)} = 1$ but $|f'| < 1$ on $\mathbb R$ (example: $f(x) = \ln(\textrm{e}^x+1)$ for $x \in \mathbb R$) is a counterexample.
A natural way to prove such a claim is to argue by contradiction and assume that there are $(x_k)_{k \in \mathbb N}, (y_k)_{k \in \mathbb N} \subset K$ such that
$$
\frac{|f(x_k) – f(y_k)|}{\|x_k-y_k\|} \le 1 – \frac{1}{k}
\qquad \text{for all } k \in \mathbb N.
$$
As $K$ is compact, after switching to a subsequence we may assume that $x_k \to x$ and $y_k \to y$ as $k \to \infty$ for some $x, y \in K$. However, I fail to see how to arrive at a contradiction if $x=y$.
Best Answer
$L$ fails to exist even in very simple cases.
Let $X=\mathbb R, K=[0,1]$ and $f(x)=\sin x$. Let $x <y$. Then $|f(x)-f(y)| =|\int_x^{y} \cos t dt| \leq |x-y|$ and equality cannot hold unless $|\cos t|=1$ for all $t \in (x,y)$. But this is is not true, so $|f(x)-f(y)| <|x-y|$.
Suppose ther exists $L <1$ such that $|f(x)-f(y)| =L |x-y|$ for al $x,y$. Then $\sin x\leq L x$ for all $x$ and we get a contradiction by dividing by $x$ and letting $x \to 0$.