Prove or disprove: function is uniformly continuous

Question

Let $f$ be a function on $\mathbb R$. If $$|f(x)-f(y)|<4|x-y|$$ for all real numbers $x$ and $y$, then $f$ is uniformly continuous on $\mathbb R$.

My answer is as follows:

The definition of Lipschitz function is: If $A⊆R$ and $f:A→R$, then $f$ is said to be Lipschitz on $A$ if there exists a constant $K>0$ such that $$|f(x)−f(y)|≤K|x−y||f(x)−f(y)|≤K|x−y|,$$

for all $x,y∈A.$

So $f$ is a Lipchitz function and every Lipchitz function is uniformly continuous function. Is it true?

Answer 1

Given $\epsilon > 0$, we want to find $\delta > 0$ such that $$ |f(x) - f(y)| < \epsilon \quad \text{whenever} \quad |x-y| < \delta $$ and $\delta$ is independent of $x$ and $y$. By Lipschitzness, if $|f(x) - f(y)| < L|x-y|$, we can define $\delta = \epsilon/L$, so the uniform continuity criterion will hold, as $L$ is independent of $x$ and $y$.