Suppose $f$ is differentiable on $(0, \infty)$ and $\lim_{x \to \infty} f'(x) = 0$.

I want to prove that $f$ is uniformly continuous on $[1, \infty)$.

So suppose the hypothesis so we have $\lim_{x \to \infty} f'(x) = 0$. Now there must exist $M$ such that $x > M$ implies $|f'(x)| < \epsilon$ (I haven't decided yet on this specific $\epsilon$). If $x > M$, then by the MVT, $\exists y, M< y < x$, such that $\frac{f(x) - f(M)}{x - M} = f'(y) < \epsilon$ which implies that $f(x) < (x- M)\epsilon + f(M).$

How should I introduce $y \in [1, \infty)$, or more specifically $f(y)$ in the above inequality?

The desired statement is for every $\epsilon > 0$, there is a $\delta > 0$ such that $|x-y| < \delta$ and $x ,y \in [1, \infty)$ implies $|f(x) - f(y)| < \epsilon$.

Also does $\lim_{x \to \infty} f(x)$ need not be finite here?

  • 1
    $\begingroup$ You have a good start. Choose $\varepsilon=1$, say. Then choose $M$ accordingly. On $[1,M]$ $f$ is uniformly continuous (why?), and on $[M,\infty)$ you have $|f(x)-f(y)|\leq |x-y|$ for every $x,y$ (why?) so $f$ is uniformly continuous there, too. $\endgroup$ – uniquesolution Oct 11 '15 at 0:44
  • $\begingroup$ Simplifying your argument a little bit, you can almost immediately check that $f'(x)$ is bounded on $[1, \infty)$. Then what can you say about $|f(x) - f(y)|$ for $x, y \in [1, \infty)$? $\endgroup$ – Sangchul Lee Oct 11 '15 at 0:51
  • $\begingroup$ @SangchulLee How do we know that $f'(x)$ is bounded on $[1, \infty)$? $\endgroup$ – user247618 Oct 11 '15 at 1:43
  • $\begingroup$ No, $f'$ need not be bounded on $[1,\infty).$ $f$ could equal $(x-1)^2\sin (1/(x-1)^2)$ on $(1,2)$ for example. $\endgroup$ – zhw. Oct 11 '15 at 1:43
  • $\begingroup$ Oh, you guys are correct. My intuition seems somehow based on locally Lipschitz functions, and obviously we need not have that nice property. Thank you for pointing out! $\endgroup$ – Sangchul Lee Oct 11 '15 at 1:46

Choose $b>1$ such that $|f'|<1$ on $[b,\infty).$ Let $\epsilon>0.$ Now $f$ is differentiable on $(0,\infty),$ hence $f$ is continuous on $(0,\infty).$ In particular, $f$ is continuous on $[1,2b],$ a compact set. Therefore $f$ is uniformly continuous on $[1,2b].$ Hence there exists $\delta > 0$ such that $|f(y)-f(x)| < \epsilon$ for $x,y \in [1,2b],|y-x| < \delta.$ Let $d'= \min(\delta,1,\epsilon).$ We know we're OK if $x,y \in [1,2b],|y-x| < \delta'.$ If $|y-x|< \delta'$ and one of $x,y$ is greater than $2b,$ then both $x,y \in [b,\infty).$ Hence by the MVT, $|f(y) - f(x)| = |f'(c)(y-x)| < 1\cdot|y-x| < \epsilon.$ Thus in all cases, if $x,y\in [1,\infty)$ and $|y-x|< \delta',$ then $|f(y)-f(x)|< \epsilon.$

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