# Suppose that $f''(x)>a$ for all $x\in \mathbb{R}$ and some $a>0$. Proof that $f$ has a absolute minimun.

Let $$f: \mathbb{R} \to \mathbb{R}$$ of class $$C^2$$ i.e $$f$$ is differentiable with continuous second derivative. Suppose that $$f''(x)>a$$ for all $$x\in \mathbb{R}$$ and some $$a>0$$. Proof that $$f$$ has a absolute minimun.

I have tried to solve this problem without any success, my attempt is:

Let $$x>0$$. By fundamental theorem of calculus we have that $$\int_{0}^{x} f''(t)dt=f'(x)-f'(0).$$ Then $$f'(x)=f'(0)+\int_{0}^{x} f''(t)dt\geq f'(0)+ax$$ for all $$x>0$$.

On the other hand, note that $$\int_{0}^{x} f'(t)dt=f(x)-f(0)$$ so $$f(x)=f(0)+\int_{0}^{x} f'(t)dt\geq f(0)+\int_{0}^{x} (f'(0)+at)dt$$ I just conclude

$$f(x)\geq f(0)+f'(0)x+\frac{1}{2}ax^2 \text{ for all } x\in \mathbb{R}$$

I have only found that $$f$$ is above a quadratic function but I would actually think that the local minimum would be the vertex of the quadratic function.

Any suggestion is appreciated.

• In your solution the continuity of $f''$ is required. The crucial inequality follows from the MacLaurin formula for $n=2,$ where $f$ is twice differentiable. Feb 23 at 17:07

Your inequality shows that the function $$f$$ tends to $$+\infty$$ both for $$x$$ going to $$-\infty$$ and to $$+\infty$$.

A continuous function that has $$+\infty$$ as its limit both for $$x$$ going to $$-\infty$$ and to $$+\infty$$ has a minimum.

This follows from the fact that there is a closed interval $$[-A,A]$$ outside of which $$f$$ is larger than $$f(0)$$. Then use the fact that a continuous function on $$[-A,A]$$ has always a minimum.

• Why vote down this? I think his answer is right.
– ZAF
Feb 23 at 16:29
• @coudy that's right! this link compliment you say math.stackexchange.com/questions/250827/… Feb 23 at 16:48

Just observe that $$f'(x)$$ is strictly increasing and hence it either tends to a limit or to $$\infty$$ as $$x\to \infty$$. If it tends to a limit $$L$$ then $$f'(x+1)-f'(x)\to L-L=0$$ By mean value theorem the left hand side equals $$f''(c)$$ and it thus always exceeds a positive number $$a$$. So the above equation can't hold and thus $$f' (x) \to\infty$$ as $$x\to\infty$$.

Similarly $$f'(x) \to-\infty$$ as $$x\to-\infty$$. And by intermediate value theorem we see that $$f'$$ vanishes somewhere say at $$c$$. Since the derivative $$f'$$ is strictly increasing, $$f'$$ vanishes only at $$c$$. Then $$f'<0$$ in $$(-\infty, c)$$ and $$f'>0$$ in $$(c, \infty)$$. Then $$f$$ is strictly decreasing in $$(-\infty, c]$$ and strictly increasing in $$[c, \infty)$$. Then $$f$$ attains absolute minimum at $$c$$.

There is no need to assume continuity of second derivative.