Real-valued, differentiable function on $\mathbb{R}$ with the bounded derivative is uniformly continuous.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable function such that there is $M>0$ and $|f'(x)|<M$ for all $x\in\mathbb{R}$.

From Mean Value Theorem we know that for each $x,y$ there is some $c$ between them such that $|f(x)-f(y)|=|f'(c)(x-y)|$. Since $f'$ is bounded by $M$, it means that $|f(x)-f(y)|<M|x-y|$. Now fix $\epsilon>0$. If we set $\delta=\frac{\epsilon}{M}$, we have $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. This holds for any $x,y \in\mathbb{R}$, thus we are done.

  • $\begingroup$ And what would the question be? $\endgroup$ Nov 6 '15 at 14:29
  • $\begingroup$ If my "proof" is correct? Thank you $\endgroup$
    – luka5z
    Nov 6 '15 at 14:30
  • 1
    $\begingroup$ Your proof is correct and you don't even need strict equalities. $\endgroup$ Nov 6 '15 at 14:45

The proof is correct, except that it would be better to write $$|f(x)-f(y)|\le M|x-y|$$ because the strict inequality fails when $x=y$.

Generally, it is preferable to use non-strict inequalities unless a strict one is needed. They are more robust: can be taken to a limit, or multiplied by $0$.


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