# If $f$ is differentiable and $f'$ is continuous on the same domain, show that for any closed subinterval, $f$ is uniformly differentiable.

Here is the problem more explicitly:

Let $$f:(a,b)\to \mathbb{R}$$ be differentiable and $$f'(a,b)\to \mathbb{R}$$ be continuous. Prove that on any closed subinterval $$[c,d]\subset (a,b)$$, the function is \textit{uniformly differentiable}, in the sense that given any $$\epsilon > 0$$, there exists a $$\delta$$ such that $$|f(y) - f(x) - f'(x)(y-x)| < \epsilon$$ for any $$x,y\in [c,d]$$ such that $$|y-x| < \delta$$.

I know that if $$f$$ is differentiable, then it is continuous. Also, the sum or product of continuous functions results in a continuous function. So I was thinking of combining them together into one function $$h:[c,d]\to \mathbb{R}$$:

$$h(x) = f(x) + (x-y)f'(x),\ \ x,y\in [c,d]$$

$$h$$ is continuous, so for any $$\epsilon > 0$$, $$\exists \delta > 0$$ such that $$|x-y| < \delta \Rightarrow |f(x) - f(y) + (x-y)f'(x) - (x-y)f'(y)| < \epsilon$$

If I could get rid of that $$f'(y)$$ term with a triangle inequality, this might work, but I can't seem to squeeze anything useful out. Perhaps due to looking at it too long or overthinking it.

Does this seem like a legitimate way to do the proof?

Firstly, $$f:[c,d]\to \mathbb{R}$$ is a continuous function on a closed and bounded (ie compact) set, so it is uniformly continuous on this set. Similarly, $$f':[c,d]\to \mathbb{R}$$ is a continuous function on a closed and bounded set, so it is bounded. Say $$|f'(x)|\leq M$$ for all $$x\in [c,d]$$. Let $$\epsilon>0$$ and let $$\delta>0$$ be such that whenever $$|x-y|<\delta$$ (with $$x,y\in [c,d]$$), then $$|f(x)-f(y)|<\epsilon/2$$. Without loss of generality, we can require $$\delta\leq \epsilon/2M$$. Then whenever $$x,y\in [c,d]$$ with $$|x-y|<\delta$$, we have $$|f(x)-f(y)-f'(x)(y-x)|\leq |f(x)-f(y)|+|f'(x)||y-x|<\epsilon/2+M\delta\leq \epsilon/2+\epsilon/2=\epsilon.$$
• The problem with your solution lies in your definition of $h$. You could try letting $h_x(y)=f(y)-f'(x)y$ (for some fixed $x$). The problem here is that you need to have this work for each $h_x$ simultaneously. So you have to use the fact that things are uniformly continuous anyway...at that point, you might as well not bother with $h$ and just prove it directly Nov 23 '19 at 22:44
Note that $$f'$$ is uniformly continuous on $$[c,d]$$, then given $$\epsilon>0$$, for some $$\delta>0$$ we have $$u,v\in[c,d]$$ and $$|u-v|<\delta$$ that $$|f'(u)-f'(v)|<\epsilon$$.
Now \begin{align*} |f(y)-f(x)-f'(x)(y-x)|&=|f'(\xi_{x,y})-f'(x)|\cdot|y-x|\\ &<\epsilon\cdot(d-c) \end{align*} since $$|\xi_{x,y}-x|\leq|x-y|<\delta$$, where $$\xi_{x,y}$$ is taken by Mean Value Theorem.