Let $f$ be uniformly differentiable if for all $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x, y \in \mathbb R$, $$\left \lvert \frac {f(x)-f(y)}{x-y} - f'(y) \right \rvert < \varepsilon $$ when $|x-y| < \delta$. Prove that if $f$ is uniformly differentiable, then $f'$ is continuous.

Note: Proofs are available; this question is to verify and critique this proof.

Proof: Suppose that $f$ is uniformly differentiable. Observe that for all $|x-y| < \delta$, the symmetric inequality $$\left \lvert \frac {f(x)-f(y)}{x-y} - f'(x) \right \rvert < \varepsilon $$ also holds, since $\frac {f(x)-f(y)}{x-y} = \frac {f(y)-f(x)}{y-x}.$ Fix an arbitrary $x \in \mathbb R$ and choose $\delta$ such that

$$\left \lvert \frac {f(x+h)-f(x)}{h} - f'(x) \right \rvert < \frac \varepsilon 3\\ $$ $$ \left \lvert \frac {f(x)-f(x-h)}{h} - f'(x) \right \rvert < \frac \varepsilon 3 $$ for $|h| < \delta$. Then, by the triangle inequality, $$ \left \lvert \frac {f(x+h)-f(x)}{h} - \frac{f(x)-f(x-h)}{h} \right \rvert < \frac 2 3 \varepsilon $$ for all $|h| < \delta$.

This in turn implies that $$ | f'(x+h) - f'(x) | \leq \frac 2 3 \varepsilon < \varepsilon $$ for all $|h| < \delta$, since if $|a_n - b_n| < k$ for all $n$, $|\lim a_n - \lim b_n| \leq k$. This proves that $f'$ is continuous.

Questions: Is this proof correct, rigorous, and well-written? Did I justify each step sufficiently ? How can it be improved?

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    $\begingroup$ Too complicated. If $\left \lvert \frac {f(x)-f(y)}{x-y} - f'(y) \right \rvert < \varepsilon$ and $\left \lvert \frac {f(x)-f(y)}{x-y} - f'(x) \right \rvert < \varepsilon$ then $|f'(x) - f'(y)| < 2 \varepsilon$. $\endgroup$
    – Martin R
    Dec 20, 2022 at 20:18

1 Answer 1


All of your steps look good to me except your "This in turn implies that..." sentence. You've written $$\left \lvert \frac {f(x+h)-f(x)}{h} - \frac{f(x)-f(x-h)}{h} \right \rvert < \frac 2 3 \varepsilon$$ and then you're trying to somehow pass to a limit to conclude $$| f'(x+h) - f'(x) | \leq \frac 2 3 \varepsilon < \varepsilon$$ but you aren't being rigorous here. What limit are you taking? If you take the limit of $\frac {f(x+h)-f(x)}{h} - \frac{f(x)-f(x-h)}{h}$ as $h \to 0$ then you'll just get $f'(x) - f'(x)$ which is $0$.

The intuition behind your proof strategy is fine, but right now you're trying to take a limit where "some of the $h$s get limited away and others remain", which is not a valid operation. We can save the proof by introducing a new variable which I'll call $s$. Then for $|h|, |s| < \frac{\delta}{2}$ you can use your same argument to get $$\left \lvert \frac {f(x+h)-f(x)}{h} - \frac{f(x-s+h)-f(x-s)}{h} \right \rvert < \frac 2 3 \varepsilon$$ and now you can take the limit as $h \to 0$ to get $$| f'(x) - f'(x-s) | \leq \frac 2 3 \varepsilon < \varepsilon.$$ Then you can finish the proof the same way you're currently doing.

Does that make sense? Basically my complaint is "you can't take a limit as some of your $h$s go to 0 while others stick around", and in general that mistake could lead you to "proofs" of totally false statements, but in this case it's easy to fix just by making the $h$s that shouldn't get limitified into a separate variable.

Side note: Martin R's comment is how I would have proved this statement. If you're handing this in somewhere then I'd recommend 1) understanding his idea, then 2) still fixing+using your own proof instead.

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    $\begingroup$ Got it, you've made clear both the problem (taking the limit of $h$ in one part of the expression and keeping it intact for the other) and the solution (keep $h$ intact but introduce a new variable $s$ and take its limit). Re "Side note: Martin R's comment" - his approach is clearly superior. Surprisingly, after reviewing his comment, I looked through my notes from exploring this problem, and saw I had at one point scrawled out $|f'(x+h) - f'(x)| \leq |\frac{f(x+h)-f(x)}h - f'(x)| + |\frac{f(x+h)-f(x)}h - f'(x+h)|$ - but seemed to forget about this when I worked on proving the symmetry. $\endgroup$ Dec 21, 2022 at 22:10
  • $\begingroup$ Actually, reviewing this further, I'm not sure I understand the problem. To me it seems I avoided this pitfall, because I didn't take the limit of $|\frac {f(x+h)-f(x)}{h} - \frac{f(x)-f(x-h)}{h}|$ but rather said that that quantity is less than $\frac 2 3 \varepsilon$ for particular $h$, and therefore $| f'(x+h) - f'(x) |$ is also, because $|a_n - b_n| \leq k \implies |\lim a_n - \lim b_n| \leq k$ (and $f'$ is a limit of $\frac {f(x+h)-f(x)}{h}$ just like $\lim a_n$ is a limit of $a_n$). My writing is certainly unclear - but is it incorrect? $\endgroup$ Dec 21, 2022 at 23:12
  • $\begingroup$ @SRobertJames When you say $|a_n - b_n| \leq k \implies |\lim a_n - \lim b_n| \leq k$, can you try stating more carefully exactly what limit(s) you're taking in that step? I think that will make it clearer. It might be a little annoying trying to explain in comments; instead I'd suggest you could edit it into a new section at the bottom of your question and then re-tag me here to let me know it's done. $\endgroup$ Dec 21, 2022 at 23:17
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    $\begingroup$ Got it. I tried doing this and saw that, as you said, I was using $h$ in two ways: In $a_n$, $h$ was fixed, and the sequence $a_n$ took the limit as $x \to x+h$, whereas, in $b_n$, the limit was as $h \to 0$ so $x+h \to x$. Thank you for clarifying (and for the suggestion of "stating more carefully exactly what limit(s) you're taking". $\endgroup$ Dec 22, 2022 at 2:14

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