Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that if $g: \mathbb{R} \rightarrow \mathbb{R}$ is twice continuously differentiable then, given $\epsilon > 0$, we can find some constant $L$ and $\delta (\epsilon) >0$ such that:

$$|g(t) - g(\alpha) - g'(\alpha)(t-\alpha)| \leq L|t-\alpha|^{2}$$

for all $|t-\alpha| < \delta(\epsilon)$.

This seems to be begging for the use of the definition of continuity on the second derivative and then somehow applying the definition of the derivative but I can't make any progress.

I can get the LHS of the inequality by using the fact that

$$g'(t) = \frac{g(t) - g(\alpha)}{t-\alpha} + \frac{o(t - \alpha)}{t-\alpha}$$

but I can't get this into any sort of inequality and besides it doesn't make use of the continuity of $g''(x)$. I've also tried to use the mean value theorem but this didn't seem to help much either. Any hints would be greatly appreciated, cheers.

share|cite|improve this question
up vote 0 down vote accepted

By the mean-value theorem $g(t) - g(\alpha) = g'(\beta)(t - \alpha)$ for some $\beta$ between $t$ and $\alpha$, so you're really trying to bound $$(g'(\beta) - g'(\alpha))(t - \alpha)$$ It should be easy from here.

You can actually get a bound of $C(t - \alpha)^2$ by applying the mean value theorem again, this time on $g'(x)$.

share|cite|improve this answer
Oops, the question was meant to say $|t-\alpha|^{2}$ but nevertheless thanks for the hint as it meant I could do the question! How did I not see this before?! – JosephML Apr 4 '14 at 15:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.