Prove if $f : [a,\infty) \longrightarrow \mathbb R$ has bounded derivative on its domain, then it is uniformly continuous.

Question

How do you prove that if $f : [a,\infty) \longrightarrow \mathbb R$ has bounded (not necessarily continuous) derivative on its domain, then it is uniformly continuous? Can you prove the same for $f : (a,\infty)$?

Some (possibly incorrect) ideas I managed to find online:

Given $f: [a, \infty) \rightarrow \mathbb{R}$ with a bounded derivative, say $|f'(x)| \leq M$ for all $x \geq a$, where $M$ is a positive constant.

Let $\epsilon > 0$ be given. We want to show that there exists a $\delta > 0$ such that for all $x, y \in [a, \infty)$ with $|x - y| < \delta$, we have $|f(x) - f(y)| < \epsilon$.

By the Mean Value Theorem, for any $x, y \in [a, \infty)$ with $x > y$, there exists $c$ between $x$ and $y$ such that $f(x) - f(y) = f'(c)(x - y)$

Since $|f'(c)| \leq M$ for all $c \geq a$, we have $|f(x) - f(y)| \leq M|x - y|$

Choose $\delta = \frac{\epsilon}{M}$. Then for any $x, y$ with $|x - y| < \delta$, $|f(x) - f(y)| \leq M|x - y| < M\frac{\epsilon}{M} = \epsilon$

Thus, we have shown that for any $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x, y \in [a, \infty)$ with $|x - y| < \delta$, we have $|f(x) - f(y)| < \epsilon$. Therefore, $f$ is uniformly continuous on $[a, \infty)$.

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I am not really sure how do we know that $|f'(c)| \leq M$ for all $c \geq a$. And how does it follow that $|x-y|<\delta,|f(x)-f(y)|\leq M|x-y|<M\frac\epsilon M=\epsilon$, assuming it's correct?

Answer 1

Its idea and using of the the Mean Value Theorem is correct, It had some wrong assumptions in its solution. Here is the right process:

According to our assumptions $f$ has bounded derivative in its domain, And it's mean:

$\forall x \in [a,\infty) , \exists M,m \in \mathbb{R}: m \le f'(x)\le M $

So According to Mean Value Theorem, We will have:

$\exists c \in [a,\infty): f(x)−f(y)=f′(c)(x−y)$

we can make an inequality for approaching to the definition of Uniform Continuous function and reaching to the conclusion with substituting $f'(c)$ with $M$ or $m$; If $(x-y)$ is positive, We will use an positive upper bound Such $M$ and make the inequality, Else if $(x-y)$ is negative, We will use an negative lower bound for inequality (for making absolute value), and in this example, we assume $(x-y)$ is positive without any interference in generality :

$\forall x,y \in [a,\infty), \exists M \in \mathbb{R}^+: |f(x)−f(y)|\le M|(x−y)|$

Then if we assign $δ = \frac{ϵ}{M}$, then we have proven the conclusion.