Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I don't know how to do this problem, formally. Let $f,g$ be real functions such that $f(a) = g(a)$, $f'(a) = g'(a)$ and for every $x$, $$ f\left( x \right) - g\left( x \right) \geq 0 $$

Prove that $f''(a) - g''(a) \geq 0$.

share|improve this question
add comment

3 Answers

up vote 3 down vote accepted

This is not true in general. Consider $a=0$ and $h(x)=f(x)-g(x)$ given for example by

$$h(x)=\begin{cases}c_1x^2&x\lt0\\c_2x^2&x\ge0\end{cases}$$

with $0\lt c_i\in\mathbb R$ and $c_1\ne c_2$, or by

$$h(x)=\begin{cases}|x^3|(1+\sin \frac1x)&x\ne0\\0&x=0\;.\end{cases}$$ Then $h(a)=0$, $h'(a)=0$, and $h(x)\ge0$, but $h$ is not twice differentiable at $a$.

The claim is true, however, if you add the assumption that $f-g$ is twice differentiable at $a$. In that case, let $h=f-g$. Then you have $h(a)=0$, $h'(a)=0$ and $h(x)\ge0$. If $h''(a)\lt0$, by the second derivative test $h$ would have to have a strict maximum at $a$, which would contradict the fact that $h(x)\ge0=h(a)$. Thus $h''(a)\ge0$, that is, $f''(a)-g''(a)\ge0$.

share|improve this answer
add comment

I assume $f,g \in C^2$. You don't mention anything about this in your problem.

Define $h=f-g$. Then by your hypothesis $a$ is a minimum point for $h$. If you assume that $h''(a)<0$ then $h'$ is decreasing in a neighborhood of $a$(which means that $h'$ is negative in a right-neighborhood of $a$) and therefore $h$ is decreasing a little bit after $a$. This contradicts the minimality of $a$.

You could fill in the blanks with some $\varepsilon$s and you're done. :)

As a note, I don't think you need the fact that $f'(a)=g'(a)$. It follows from the fact that $f-g \geq 0$ and $(f-g)(a)=0$.

share|improve this answer
add comment

If $(f-g)''$ is defined locally, just apply the definition of the second derivative for a function $h$ and look at the limit at $a$ of $\frac{h(x)-h(a)-h'(a)(x-a)}{(x-a)^2}$. Here $h=f-g$ so $h(a)=0$ and $h'(a)=0$ and $h(x)\geq 0$ so the limit is positive right ?

share|improve this answer
add comment

Your Answer

 
discard

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.