Let $E=[0,1]$, $K>0$ be a given constant, and $F$ be a set of continous functions on compact interval E such that $|f(x)-f(y)|\leq K|x-y|$

Question

Let $E=[0,1]$, $K>0$ be a given constant, and $F$ be a set of continous functions on compact interval E such that $|f(x)-f(y)|\leq K|x-y|, x,y\in E, f\in F$ $\\$, and $f(0)=0$.

Show that $F$ is uniformly bounded and equicontinuous.

My approach is that let $\delta = \frac \epsilon K$, then $|f(x)-f(y)|\leq K|x-y|\le \epsilon$ when $|x-y|\le \delta$.

This would finish the proof for equicontinuous.

Also, we know each $f$ is pointwise bounded, which implies $F$ is uniformly bounded.

Is my approach correct? And why the condition $f(0)=0$ is never used?

Answer 1

That is for uniformly boundedness: $|f(x)|=|f(x)-f(0)|\leq K|x-0|=K|x|\leq K$ for $x\in[0,1]$ and $f\in F$.

Answer 2

If you don't have that $f(0)=0$ then all constant functions are in $F$, so your space wouldn't be uniformly bounded.