# Prove that $f$ is uniformly continuous

I have to prove this:

Suppose $$f:(a,b)\to \mathbb{R}$$ is differentiable and $$|f'(x)| \leq M$$ for all $$x\in (a,b)$$. Prove that $$f$$ is uniformly continuous on $$(a,b)$$.Give an example of a function $$f:(0,1) \to \mathbb{R}$$ that is differentiable and uniformly continuous on $$(0,1)$$, but such that $$f'$$ is unbounded.

My attempt:

Since $$f$$ is differentiable we know that for all $$\epsilon>0$$ there exists a $$\delta>0$$ such that if $$|x-y|<\delta$$ then:

$$\left|\frac{f(x)-f(y)}{x-y}-f'(x)\right|< \epsilon$$

then we choose $$\delta=\frac{\epsilon}{\epsilon+M}$$ and we have that:

$$\begin{multline} \left|\frac{f(x)-f(y)}{x-y}-\right| \epsilon+f'(x) \leq \epsilon +M \\ \Rightarrow |f(x)-f(y)|<(\epsilon +M)|x-y|<(\epsilon +M) \delta= \epsilon \end{multline}$$

Therefore $$f$$ is uniformly continuous.

Now let's consider the function $$f: \mathbb{R_{\geq 0}} \to \mathbb{R}$$ such that $$f(x)= \sqrt{x}$$. To see that $$f$$ is uniformly continuous we note that it is continuous at all its dominion, therefore $$f$$ is continuous at $$[0,1]$$ then we have that $$f$$ is uniformly continuous at $$[0,1]$$, this means that the same $$\delta$$ Works for $$x,y \in (0,1)$$, and therefore $$f$$ is uniformly continuous at $$(0,1)$$ , but $$f'(x)=1/2 \sqrt{x}$$ that is clearly unbounded.

Can you tell me if I am right please, if not how can I fix the problems please, thank you a lot.

• Correct, in the first part you essentially prove that if $|f'|\leq M$ then $f$ is $M$-Lipschitz, which implies uniform continuity. Second part correct. Oct 19, 2014 at 3:23
• Thanks Milly then I don't have to fix something, right? :) Oct 19, 2014 at 3:26
• It is all right, apart from the typo in formula after "then we choose $\delta$ ..." (I think meant $\frac{|f(x)-f(y)|}{|x-y|}\leq \left|\frac{f(x)-f(y)}{x-y}-f'(x)\right|+|f'(x)|\leq \varepsilon+M$). Oct 19, 2014 at 3:29

Example is $f(x)=x^{\frac{1}{n}}$, basically you have to construct a function whose limit exists at end points and derivative is unbounded in $(0,1)$

You write ‘Since $$f$$ is differentiable we know that for all $$\epsilon > 0$$ there exists a $$\delta > 0$$’, but you don't know that unless $$f$$ is uniformly differentiable (which is a stronger condition that doesn't follow from what you're given). Then in the next line you say that $$\delta$$ is $$\epsilon / ( \epsilon + M )$$, but even if $$f$$ were uniformly continuous, you have no way of knowing that $$\delta$$ can be that large.

Now, you are right to try to say that, given $$\epsilon$$, you should find $$\delta$$, because you're trying to prove that $$f$$ is uniformly continuous, and that's how that goes. So you do want to prove that $$\lvert f ( x ) - f ( y ) \rvert < \epsilon$$ whenever $$\lvert x - y \rvert < \delta$$, but you can't assume that $$\big \lvert \big ( f ( x ) - f ( y ) \big ) / ( x - y ) - f ' ( x ) \big \rvert < \epsilon$$ along the way. So I won't use that.

Whatever intuition led you to $$\delta = \epsilon / ( \epsilon + M )$$ is sound. You can basically draw a picture of the situation to guess that $$\delta = \epsilon / M$$, but since it doesn't do to cut things too close, you should make $$\delta$$ slightly smaller than this (while keeping it positive), and $$\epsilon / ( \epsilon + M )$$ does the trick. As it happens, in this case, you can simply take $$\delta$$ to be $$\epsilon / M$$, so that is what I'll do here, but it doesn't hurt to give yourself some breathing room.

Now I think that you should use the Mean Value Inequality (MVI). This says that any constant bounds satisfied by a derivative must also be satisfied by the difference quotients. In this case, since $$\lvert f ' \rvert \leq M$$, you know that $$\big \lvert \big ( f ( x ) - f ( y ) \big ) / ( x - y ) \big \rvert \leq M$$. (You can think of this as following from the Mean Value Theorem; if this bound were ever violated, then it would be violated by the derivative where it equals the difference quotient, which the MVT says must happen somewhere between $$x$$ and $$y$$. But in my opinion, the MVI is actually the more fundamental result, and the MVT should be seen as a corollary of the MVI rather than the reverse.)

Then if $$\lvert x - y \rvert < \delta$$, you immediately get $$\big \lvert f ( x ) - f ( y ) \big \rvert \leq M \lvert x - y \rvert < M \delta = M ( \epsilon / M ) = \epsilon$$.

If you'd rather use the MVT than the MVI, then you can run this backwards as a proof by contradiction. Again, the claim is that $$\delta = \epsilon / M$$ shows that $$f$$ is uniformly continuous, but let's suppose (by way of contradiction) that there exist $$x$$ and $$y$$ such that $$\lvert x - y \rvert < \delta$$ but $$\lvert f ( x ) - f ( y ) \rvert \geq \epsilon$$. Then $$\big \lvert \big ( f ( x ) - f ( y ) \big ) / ( x - y ) \big \rvert > \epsilon / \delta = M$$, which by the MVT means that $$\lvert f ' ( z ) \rvert > M$$ for some $$z$$ between $$x$$ and $$y$$, but we're given that this is false.

• I didn't address your counterexample in the second part; that is perfectly good. Apr 29, 2021 at 6:13