Lipschitz's continuity general question

Question

Prove that lipschitz continuous (in its domain) function $f(x)$ defined on a bounded set, has bounded range.

I do not know whether this is not directly implied by the definition, nevertheless I do not know what to write in this examples as well.

Answer 1

By Lipschitz and triangle inequality $$|f(x)|-|f(y)|\leq|f(x)-f(y)|<L|x-y|.$$

It follows that $$|f(x)|\leq L|x-y|+|f(y)|.$$ Now fix $y$ and move $x$.

Since the domain of $f$ is bounded, say by $M$. Then $|x|<M$. Therefore $|x-y|\leq|x|+|y|<M+|y|$. Therefore $$|f(x)|\leq L|x-y|+|f(y)|<LM+L|y|+|f(y)|.$$

This means that $|f(x)|$ is in the ball with center $0$ and radius $LM+L|y|+|f(y)|$, i.e. it is bounded.

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I wrote it in the reals to not make it obscure with unnecessary generality. On metric spaces just replace the absolute values with distances:

$$d(f(x),a)-d(f(y),a)\leq d(f(x),f(y))<Ld(x,y),$$ where $a$ is some fixed point. So

$$d(f(x),a)\leq Ld(x,y)+d(f(y),a).$$

Therefore $f(x)$ is in the ball with center $a$ and radius $LM+Ld(a,y)+d(f(y),a)$.