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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.

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  • $\begingroup$ I find your “proof” hard to follow and check. You should rewrite it while defining precisely all your variables and checking dependencies. $\endgroup$
    – Aphelli
    Dec 12, 2018 at 22:45
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    $\begingroup$ 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. $\endgroup$ Dec 12, 2018 at 22:48

1 Answer 1

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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!

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