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

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3 Answers 3

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


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, $f''(a)-g''(a)\ge0$.

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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 $(f-g)(a)=0$.

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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 ?

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