Show that if a function $f$ is Lipschitz continuous on $X$, $f$ has to be uniformly continuous on $X$.

**My attempt:**

**(1)** The definition of Lipschitz continuity for $f$ on $X$ is:

$exists L in mathbb{R}^+_0 ,,,forall x,y in X: |f(x)-f(y)|le L |x-y|Longleftrightarrow frac{|f(x)-f(y)|}{|x-y|}le L$

For the case $L=0$ it has to hold that $,,forall x,y in X:|f(x)-f(y)|=0$ this means the function is constant and therefore uniformly continuous.

Now let $0ne L=frac{epsilon}{delta}$ with two corresponding $epsilon,deltainmathbb{R}^+$ or in other words $forall epsilon >0,,, existsdelta>0:frac{epsilon}{delta}=L$

We now know $frac{|f(x)-f(y)|}{|x-y|}le frac{epsilon}{delta}=L$

that can only be true when $forall epsilon >0,,, existsdelta>0,,,forall x,yin X:|x-y|<deltaLongrightarrow|f(x)-f(y)|<epsilon$

Which is the Heine-Cantor definition of uniform continuity.

So if $f$ isnt uniformly continuous, than $f$ also cannot be Lipschitz continuous.

Would be great if someone could look over it!