Uniform continuity problem

Question

Let $f(x)$ be continuous on $[0, \infty)$, $f'(x)$ and $f''(x)$ be continuous on $(0, \infty)$. Which of the following statements are true:

I. If $f'(x) > 0$ and $f''(x) < 0$, then f(x) is uniformly continuous on $[0, \infty)$

II. If $f'(x) < 0$ and $f''(x) > 0$, then f(x) is uniformly continuous on $[0, \infty)$

III. If $f'(x) < 0$ and $f''(x) < 0$, then f(x) is uniformly continuous on $[0, \infty)$

I think that I and III are false.

Counterexample for I: $f(x) = \sqrt{x}$

Counterexample for III: $f(x) = -x^2$

EDIT: I deleted my "counterexample" for II because it was a mistake. So, my question is how to prove/disprove II?

[$f(x)=\sqrt x$ is uniformly continuous on $0,\infty)$. You can adapt the argument in the first answer there to show I is true (and then II is true by considering $-f$). — David Mitra, Jul 03 '14 at 18:00

score 1 · Answer 1 · answered Jul 03 '14 at 18:16

Note I and II are equivalent since you can go from one to the other by replacing $f$ by $-f$.

Focusing on I, it suffices to show uniform continuity on $[1,\infty)$ since just by continuity $f$ is uniformly continuous on $[0,1]$. If $y > x \geq 1$, then by the mean value theorem $$f(y) - f(x) = f'(z)(y - x)$$ Here $z$ is between $x$ and $y$. Since $f'' < 0$, you have $f'(z) < f'(1)$. So $$f(y) - f(x) < f'(1)(y - x)$$ Note also $f' > 0$, so the left hand side is positive and for $M = f'(1)$ you have $$|f(y) - f(x)| < M |y - x|$$ So $f$ is Lipschitz on $[1,\infty)$ which is enough for uniform continuity.

Uniform continuity problem

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