A function have minimum or maximum

Question

Let $f$ be strictly convex or concave on $\Bbb R$. $\lim_{x\to -\infty}f'(x)=A<0, \lim_{x\to+\infty}f'(x)=B<0$. Can we show that $f$ attains a global minimum or maximum?

I could not even construct such an example satisfying the assumptions.

Answer 1

If $$f(x)=-\ln(1+e^{-x})-2x$$

then $$f'(x)=-1-\frac{1}{1+e^{-x}}$$

and we have $\lim_{x\to-\infty}f'(x)=-1<0$ and $\lim_{x\to\infty}f'(x)=-2<0$. The function $f$ is strictly concave because

$$f''(x)=-\frac{e^{-x}}{(1+e^{-x})^2}<0$$

for all $x\in\mathbb{R}$.

The function $f$ has no global maximum or minimum (its derivative is always negative).

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Note that I came up with the example by taking $f'$ to be the negative of a logistic function (which is a function with finite limits at $-\infty$ and $\infty$) and subtracting one (to make both the limits negative).

(You could construct a strictly convex $f$ in a similar way, just by subtracting a larger number from the logistic function.)

Answer 2

Suppose $f'$ exists for every $x\in\mathbb R$, if $f$ is strictly convex then $f'$ is strictly crescent so we must have $A<f'(x)<B<0$ and can't have any minimum-maximum because $m$ minimum (maximum) implies $f'(m)=0$. If $f$ is concave then $f'$ is strictly decrescent then it must be $B<f'(x)<A<0$ and so on.

Answer 3

Take a generic function $f(x)$ that meets the required criteria. Then, using the definition of a limit, we can divide $f(x)$ into three intervals: $(-\infty,a)$, where $|f(x)-(C_1+Ax)|<\epsilon$; $(b,\infty)$, where $|f(x)-(C_2+Bx)|<\epsilon$; and $[a,b]$, which is a closed interval, and therefore $f(x)$ has a maximum and a minimum value within it, which we will call $M$ and $N$ respectively. The approximations for the lines both have negative slope, which means that points $a'$ and $b'$ exist such that, $\forall x<a', f(x)>M+\epsilon>M$ and $\forall x>b', f(x)<N-\epsilon<N$.

As such, there is no global maximum or minimum, although I also can't think of any examples for functions that actually meet the criteria.