Lipschitzian function implies uniformly continuity - choice of delta

Question

On this link which is french page of uniformly continuity, it is said that if a function is k-Lipschitzian, then this function is uniformly continuous.

They say that by choosing a $\delta$ as :

$$k\delta \leq \epsilon$$

Shouldn't it be rather $\delta$ as :

$$\delta \geq \epsilon/k\quad\quad?$$

Indeed, for a fixed $\epsilon$, we want to choose a $\delta$ as :

$$\dfrac{f(x+\delta)-f(x)}{\delta} = \dfrac{\epsilon}{\delta}$$

Moreover, we have with k-Lipschizian definition :

$$\dfrac{f(x+\delta)-f(x)}{\delta} \leq k$$

So we can choose $\delta$ as :

$$\dfrac{\epsilon}{\delta} \leq k$$

which is equivalent to: $\delta \geq \epsilon/k$, isn't it ?

Any help is welcome

Answer 1

You want $\left|\frac{f(x+\delta)-f(x)}{\delta} \right| \leq \frac{\epsilon}{\delta}$ or to be more precise: $$\left|\frac{f(y)-f(x)}{\delta} \right| \leq \frac{\epsilon}{\delta}, \forall y \text{ such as } |x-y| \leq \delta$$ And you have: $$\left|\frac{f(y)-f(x)}{\delta} \right| \leq \frac{k |x-y|}{\delta}\leq k, \forall y \text{ such as } |x-y| \leq \delta$$.

So you have to choose $\delta$ such has: $$a \leq k \Rightarrow a \leq \frac{\epsilon}{\delta}$$ or in term of intervals: $$(0,k] \subset \left(0,\frac{\epsilon}{\delta} \right]$$ which gives: $$k \leq \frac{\epsilon}{\delta}$$ i.e $$\delta \leq\frac{\epsilon}{k}$$

Answer 2

The definition of Lipschitz (even the French one) says that a function $f$ is Lipschitz (on some interval $E$) if

$$|f(x) - f(y)| \leq k|x-y| $$

for some $k$, for any choice of $x,y \in E$.

Now, given any $\epsilon$, if we pick $\delta \leq \epsilon/k$, then for any pair $x,y \in E$ with $|x-y| < \delta$

$$|f(x)-f(y)| \leq k|x-y| < k \frac{\epsilon}{k} = \epsilon $$

which means by definition that $f$ is uniformely continuous on $E$.

Picking any $\delta \geq \epsilon/k$ makes no sense, what if I pick a huge $\delta$?