Show that if $f\in E$, then $f$ is uniformly continuous.

Question

Here's the problem statement.

Let $E$ be the set of functions $f:[0,1]\to \mathbb{R}$ such that $K(f)=\text{sup}${$\frac{|f(x)-f(y)|}{|x-y|}: x,y\in [0,1] \text{ with } x\neq y$}$<\infty$.

(i) Show that if $f\in E$, then $f$ is uniformly continuous.

For this part, for each $x,y\in [0,1]$, $\frac{|f(x)-f(y)|}{|x-y|}\leq K(f)$, and since $|x-y|\leq 1$, $|f(x)-f(y)|\leq K(f)$. Actually, I'm a bit confused at this point.

(ii) Prove that $f,g\in E$, $\lambda, \mu \in \mathbb{R}$ implies $\lambda f+\mu g\in E$, $K(\lambda f+\mu g)\leq |\lambda|K(f)+|\mu|K(g)$ and $|K(f)-K(g)|\leq K(f-g)$.

I think the first one can be proved just based on the definition, but I have no idea for the rest.

Answer 1

1. Observe that, if $f\in E$ and $x\neq y$, $|f(x)-f(y)|\leq K(f) |x-y|$, which shows that $f$ is Lipschtitz continuous, hence uniformly continuous.
2. For all $\lambda\in \mathbb R$ and $f\in E$ we have that $$\sup_{x\neq y}\frac{|\lambda f(x)-\lambda f(y)|}{|x-y|}=|\lambda| K(f)<\infty,$$ hence $\lambda f\in E$, moreover $K(\lambda f)=|\lambda|K(f)$.

Given $f, g\in E$ and $x\neq y$, by the triangle inequality we have $$\frac{| f(x)+g(x)- f(y)-g(y)|}{|x-y|}\leq\frac{|f(x)-f(y)|}{|x-y|}+\frac{|g(x)-g(y)|}{|x-y|}.$$ The same inequality holds when taking $\sup_{x\neq y}$ on both sides:

\begin{align} \sup_{x\neq y}\frac{| f(x)+g(x)- f(y)-g(y)|}{|x-y|} \leq \sup_{x\neq y}\left(\frac{|f(x)-f(y)|}{|x-y|}+\frac{|g(x)-g(y)|}{|x-y|}\right)\\ \leq\sup_{x\neq y}\left(\frac{|f(x)-f(y)|}{|x-y|}\right)+\sup_{x\neq y}\left(\frac{|g(x)-g(y)|}{|x-y|}\right)=K(f)+K(g)<\infty. \end{align}

Combining these two facts proves that $f, g\in E$, $\lambda, \mu\in \mathbb R$ implies $\lambda F+\mu g\in E$ and $$K(\lambda F+\mu g)\leq K(\lambda f)+K(\mu g)=|\lambda|K(f)+|\mu|K(g).$$

For the last question, recall that $|a-b|\geq |a|-|b|$, so $$\frac{|f(x)-g(x)-(f(y)-g(y))|}{|x-y|}\geq\frac{|f(x)-f(y)|}{|x-y|}-\frac{|g(x)-g(y)|}{|x-y|}\geq \frac{|f(x)-f(y)|}{|x-y|}-K(g).$$ Taking the supremum on both sides yields $$K(f-g)\geq K(f)-K(g).$$

Now, by what we have shown above, $K(f-g)=K(g-f)\geq K(g)-K(f)=-(K(f)-K(g))$, hence $$ K(f-g)\geq -(K(f)-K(g))\quad \text{ and } \quad K(f-g)\geq K(f)-K(g),$$ which implies $K(f-g)\geq |K(f)-K(g)|$.