Ytukyg

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

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

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

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

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

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

for all $xinmathbb{R}$.

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

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.)

Suppose $f'$ exists for every $xinmathbb R$, 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.

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.