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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?

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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 $$ |f(x)-f(y)-f'(x)(y-x)|\leq |f(x)-f(y)|+|f'(x)||y-x|<\epsilon/2+M\delta\leq \epsilon/2+\epsilon/2=\epsilon. $$

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  • $\begingroup$ I like this solution, but what critique would you give my approach? $\endgroup$ Nov 23 '19 at 22:29
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    $\begingroup$ 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 $\endgroup$ Nov 23 '19 at 22:44
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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.

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