Let $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that $f''(x)\geq a$ for all $x \in \mathbb{R}$. Show that $f$ has an absolute minimum.

Question

Let $ f \colon \mathbb{R} \rightarrow \mathbb{R} $ twice differentiable such that $ f''(x)\geq a $ for all $ x \in \mathbb{R} $ and some $a>0$. Show that $f$ has an absolute minimum.

I'm trying to do this exercise for absurdity, but I have not been able to obtain good results, for this reason I refrain from uploading my failed results. I would like someone to please help me to demonstrate this. I appreciate any suggestions you can give me.

Answer 1

If $f''(x)$ is strictly positive for each $x$, then $f'$ is a strictly increasing function. This means that there is at most one point $a$ such that $f'(a) = 0$. Moreover, if such a point exists, then $f'(x) < 0$ for $x < a$ and $f'(x) > 0$ for $x > a$. In other words, $f$ is decreasing on the interval $(\infty, a]$ and increasing on $[a, \infty)$, meaning that such an $a$ is indeed a global minimum.

Now, does such an $a \in \mathbb{R}$ exist? Pick any $b \in \mathbb{R}$. If you're very lucky and $f'(b) = 0$, you are done. Otherwise, observe that $f'(b+x) \geq f'(b) + ax$ and $f'(b-x) \leq f'(b) - ax$ for $x > 0$, since $f''(x) \geq a$ for each $x \in \mathbb{R}$. Thus, if $f'(b) < 0$, then $f'(b+x) > 0$ for large enough $x$, hence by the Intermediate Value Theorem $f'$ takes the value $0$ somewhere to the right of $b$. On the other hand, if $f'(b) > 0$, then $f'(b-x) < 0$ for large enough $x$, hence by the Intermediate Value Theorem $f'$ takes the value $0$ somewhere to the left of $b$.

Answer 2

Let's first prove that there is a real number $c$ such that $f'(c)=0$.

Suppose there isn't, and because $f'$ is differentiable, it is continuous, hence either $f'(x)$ is always positive or always negative.

If it is always positive, then $f'(0)$ and $f'(-f'(0)/a)$ are both positive. Consider that

$$f'(0)-f'(-f'(0)/a)=\int_{-f'(0)/a}^0 f''(x)dx$$ $$f'(0)-f'(-f'(0)/a)\ge\int_{-f'(0)/a}^0 a\ dx$$ $$f'(0)-f'(-f'(0)/a)\ge f'(0)$$ $$f'(-f'(0)/a)\le 0$$

Contradiction. Therefore it must be always negative, so $f'(0)$ and $f'(-f'(0)/a)$ are both negative. Now consider that

$$f'(-f'(0)/a)-f'(0)=\int_{0}^{-f'(0)/a} f''(x)dx$$ $$f'(-f'(0)/a)-f'(0)\ge\int_{0}^{-f'(0)/a} a\ dx$$ $$f'(-f'(0)/a)-f'(0)\ge -f'(0)$$ $$f'(-f'(0)/a)\ge 0$$

Another contradiction. Therefore the only possibility is that there is a real number $c$ such that $f'(c)=0$.

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Now because $f''(x)$ is always positive, it must be that $f'(x)<0$ for $x<c$ and $f'(x)>0$ for $x>c$. Therefore there is exactly one stationary point of $f$, which is $c$, and $f$ is descending before $c$ and ascending after $c$. In conclusion, $f(c)$ is the absolute minimum of $f$, Q.E.D.