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I need to prove the following statement:

Suppose $f(x)$ is continuous on $[1, + \infty)$ and differentiable on $(1, + \infty)$ and there exists a finite limit $\lim_{x \to + \infty} f'(x)$ then $f(x)$ is uniformly continuous on $[1, + \infty]$.

If $f'(x)$ is continuous then $f'(x)$ is bounded, hence $f(x)$ is uniformly continuous. But $f'(x)$ can be discontinuous, because $f(x)$ is not continuously differentiable, so we cannot say that $f'(x)$ is bounded. OK, If $f'(x)$ has removable or jump discontinuity then it doesn't effect the boundness (am I right?) and we still can say that $f'(x)$ is bounded, therefore $f(x)$ is uniformly continuous.

But if $f'(x)$ has an essential discontinuity, must $f'(x)$ be neccessary bounded in this case also, or should I find another way of proving the unifom continuity of $f(x)$?

Thanks a lot in advance for any help!

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  • $\begingroup$ "If $f(x)$ has removable or jump discontinuity then" $f$ isn't differentiable on $(1, \infty)$. $\endgroup$ – Arthur Apr 10 '18 at 11:29
  • $\begingroup$ @Arthur So, if the derivative of the differentiable function is discontinuous then it can only have essential discontinuities? $\endgroup$ – D F Apr 10 '18 at 11:46
  • $\begingroup$ Not directly related to what I said. I said that if a function is discontinuous, then it's not differentiable. But yes, if a derivative is defined on each point on an interval, then that derivative cannot have a jump discontinuity. $\endgroup$ – Arthur Apr 10 '18 at 11:48
  • $\begingroup$ @Arthur oh, I've mistaken, yes thanks, I meant $f'(x)$ instead of $f(x)$ $\endgroup$ – D F Apr 10 '18 at 11:50
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If $\lim_{x \to + \infty} f'(x)=L\in\mathbb{R}$ then by the definition of limit, for $\epsilon=1$ there is $a\geq 1$ such that for all $x\geq a$, $$L-1\leq f'(x)\leq L+1$$ that is $|f'|$ is bounded by $M:=|L|+1$ in $[a,+\infty)$.

Thus, by the Mean Value Theorem (where we need only the differentiability of $f$ and we need not that $f'$ is continuous), for $a\leq x<y$, there exists $t\in(x,y)$ such that $$|f(x)-f(y)|=|f'(t)||x-y|\leq M|x-y|$$ which implies that $f$ is uniformly continuous in $[a,+\infty)$.

Moreover, by Heine-Cantor theorem, $f$ is uniformly continuous in the compact set $[1,a]$ (here we need not the differentiability of $f$).

Hence we may conclude that $f$ uniformly continuous in $[1,a]\cup [a,+\infty)=[1,+\infty)$.

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  • $\begingroup$ Thanks a lot for your answer. I know, that your first statement holds for continuous functions, but if $f'(x)$ is discontinuous, is it still true? $\endgroup$ – D F Apr 10 '18 at 11:42
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    $\begingroup$ Yes, because by Lagrange theorem, if $f'$ is bounded then $f$ is Lipschitz $\endgroup$ – Robert Z Apr 10 '18 at 11:46
  • $\begingroup$ I'm very sorry, but I still can't understand why your first statement holds for discontinuous function... $\endgroup$ – D F Apr 10 '18 at 11:56
  • $\begingroup$ What is "first statement"? I'm saying that: If $f$ is differentiable in $[a,+\infty)$ and $f'$ is bounded in $[a,+\infty)$ then $f$ is Lipschitz in $[a,+\infty)$. (even if $f'$ is not continuous). $\endgroup$ – Robert Z Apr 10 '18 at 12:00
  • $\begingroup$ I meant this "if $\lim_{x \to + \infty} f'(x)$ exists and is finite then $f'(x)$ is bounded on $[a, + \infty)$" $\endgroup$ – D F Apr 10 '18 at 12:02
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Assume there is a some $\epsilon>0$ and sequences $x_n,y_n$, such that $|x_n-y_n|\to0$ and $|f(x_n)-f(y_n)|>\epsilon$.

We can't have $x_n$ bounded, because $f$ is uniformly continuous on each compact interval. Therefore, we can assume that $x_n\to\infty$.

Now, $|f(x_n)-f(y_n)|=|f'(z_n)||x_n-y_n|$ for some $z_n$ between $x_n,y_n$.

Since $f'(x)\to L$ as $x\to\infty$ we have $f'(z_n)$ bounded. Therefore $|f(x_n)-f(y_n)|\to0$, since $|x_n-y_n|\to0$. Contradiction. Therefore, such pair of sequences do not exist. This means that $f$ is uniformly continuous.

Nota bene: $f'$ bounded doesn't follow from the existence of the limit $\lim_{x\to\infty}f'(x)$. One only gets bounded in some neighborhood of $\infty$.

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