# Prove or disprove that there exists a constant $K$ such that $|f(x)-f(y)|\leq K |x-y|,\;\;\forall\;\;x,y\in[0,1]$

I'm preparing for my final exam and I want to solve as many questions as possible. However, I don't know how to tackle this question. Please, can anyone help me out?

Let $f$ be a function on $[0,1]$ into $\Bbb{R}$. Suppose that if $x\in[0,1],$ there exists $K_x$ such that $|f(x)-f(y)|\leq K_x |x-y|,\;\;\forall\;\;y\in[0,1]$ Prove or disprove that there exists a constant $K$ such that $|f(x)-f(y)|\leq K |x-y|,\;\;\forall\;\;x,y\in[0,1]$

## 1 Answer

This is disproven by, for instance, $$f(x)=\cases{0& if x=0\\x\sin(1/x)& otherwise}$$ To elaborate, $f$ has unbounded derivative, meaning it's not Lipschitz.

However, for each $x$ there is a $K_x$. Indeed, $K_0=1$ works, and for $x\in(0,1]$, let $x_0\leq x$ be such that $f'$ has a local maximum or minimum at $x_0$. Then $K_x=|f'(x_0)|$ works.

• Thank you very much! Please, can you elaborate further? – Omojola Micheal May 20 '18 at 3:54
• @Mike I added a small elaboration. – Arthur May 20 '18 at 5:07