# Proving that a differentiable function f, with f' bounded, is uniformly continuous

Let $$f: R \rightarrow R$$ be a differentiable function. Prove that if $$f'$$ is bounded, then $$f$$ is uniformly continuous

My attempt:

Since $$f$$ is differentiable, we have $$f'(x) = \lim_{x \to x_0}\frac{f(x) - f(x_0)}{x-x_0}$$.

Since $$f$$ is differentiable, it follows that $$f$$ is also continuous. Therefore, given $$\epsilon > 0$$, there is some $$\delta > 0$$ such that $$|x-x_0|<\delta \rightarrow |f(x)-f(x_0)|<\epsilon$$.

$$\frac{|x-x_0|}{|x-x_0|}*|f(x)-f(x_0)|=|x-x_0|\frac{|f(x)-f(x_0)|}{|x-x_0|}<\delta\frac{|f(x)-f(x_0)|}{|x-x_0|}<\epsilon$$

Now take $$\lim_{x \to x_0}$$ of both sides to get: $$\delta f'(x) < \epsilon$$. Since $$f'$$ is bounded, there is some $$M$$ such that $$-M \leq f' \leq M$$. So we have $$\delta f'(x)\leq\delta M<\epsilon$$. Take $$\delta = \frac{\epsilon}{M}$$. Therefore, $$f$$ is uniformly continuous.

Is my logic here correct? I am not sure if I can do the limit step.

• I find your “proof” hard to follow and check. You should rewrite it while defining precisely all your variables and checking dependencies. Dec 12, 2018 at 22:45
• Yes. Use Lagrange's Mean Value theorem. $|f(x)-f(x_0)|=|f'(c)(x-x_0)|\leq M|x-x_0|$ where $c\in(x,x_0)$ and $Min\mathbb{R}\::\:|f'(x)|\leq M\:\forall \:x\in\mathbb{R}$. Infact it is $M$-Lipschitz. Dec 12, 2018 at 22:48

The $$\delta$$ you end up choosing is good; this $$\delta$$ will work. The problem is your justification why it works. It almost works with a bit of unpacking, but it needs to be rewritten in a way that is clear flows logically from the assumption that $$|x - x_0| < \delta = \varepsilon/M$$ to the conclusion that $$|f(x) - f(x_0)| < \varepsilon$$.

You write

Since $$f$$ is differentiable, it follows that $$f$$ is also continuous. Therefore, given $$\varepsilon > 0$$, there is some $$\delta > 0$$ such that $$|x-x_0|<\delta \rightarrow |f(x)-f(x_0)|<\varepsilon$$.

This is true, but it's a confusing step to put in your proof. The way I interepeted this when I first read it is that you are choosing a value for the variable $$\delta$$. You don't know what it is, but you know one exists that satisfies the above property (given the fixed $$\varepsilon > 0$$ and $$x_0 \in \mathbb{R}$$). The rest of your proof would then (somehow) have to justify why this $$\delta$$ does not depend on $$x_0$$ at all, or perhaps define another $$\delta'$$, based on $$\delta$$, that didn't depend on $$x_0$$. Since you end up simply defining a totally different $$\delta$$, not relating to this $$\delta$$, the step is confusing, and probably should be dropped.

Next, you write

$$\frac{|x-x_0|}{|x-x_0|} \times |f(x)-f(x_0)|=|x-x_0|\frac{|f(x)-f(x_0)|}{|x-x_0|}<\delta\frac{|f(x)-f(x_0)|}{|x-x_0|}<\varepsilon$$

This is another reason why the previous step is confusing; you are now using $$\delta$$ in inequalities. Without first defining $$\delta$$, this is a bit confusing. You haven't talked about what you're assuming here. For example, it appears that you're assuming $$|x - x_0| < \delta$$. If you have a $$\delta$$ defined (such as from the previous paragraph), this is a very reasonable assumption to make, but it should be made explicit before writing this step.

It's also not clear why $$\delta\frac{|f(x)-f(x_0)|}{|x-x_0|}<\varepsilon$$. Without a clear definition of $$\delta$$, I don't know why this would be true.

You also write

Now take $$\lim_{x \to x_0}$$ of both sides to get: $$\delta f'(x) < \epsilon$$.

This is a small point: if you have $$g(x) < N$$ for all $$x$$, then $$\lim_{x \to a} g(x) \le N$$. You cannot conclude strict inequality!