In some cases it turns out that,
If $f$ is strictly increasing, and convex then all its higher order derivatives are also strictly increasing and convex (assuming $f$ is continuously differentiable).
For example,
- $e^{a x}$ on $\mathbb{R}$ for $a>0$,
- $\tan(x)$ on $(0,\frac{\pi}{2})$ or,
- $x^n$ on $(0,\infty)$ – at least for the first few derivatives.
Of course several counter examples, eg. $x \log x$, which is convex and increasing on $(1,\infty)$ but its derivatives do not share this property.
So my question is that under what conditions does it happen that if $f$ is increasing then its derivatives are also monotone (when are they increasing or decreasing)??
By conditions I mean conditions of $f$, not its higher derivatives. I know this is somewhat unusual in that I am trying to infer information about $f'$ knowing information about $f$. In most calculus courses the reverse is deduced.
Thanks for reading
Best Answer
One property that will give us $f'$ strictly increasing is the following : we say that $f$ is convex if $$ \forall \lambda \in [0,1], \quad f(x + \lambda( y-x)) \le f(x) + \lambda( f(y) - f(x) ) $$ and we say that $f$ is strictly convex if the inequality is always strict. (The intuition behind this is that if you draw the cord between the points $(x,f(x))$ and $(y,f(y))$, the cord cannot touch the graph, thus we cannot have an interval where the derivative is constant if $f$ is strictly increasing.)
Assume $f$ is strictly convex and differentiable. Then $f'$ is strictly increasing. Since strictly convex functions are convex, we already know that $f'$ is increasing. Suppose there exists $x \le y$ such that $f'(x) = f'(y)$. Then this means that in the interval $[x,y]$, the function $f'$ is constant. But then $f$ cannot be strictly increasing on this interval unless $x=y$, thus $f'$ must be strictly increasing. The reason for this is that strict convexity on $[x,y]$ with $x < y$ implies that by the mean value theorem, there exists $c_1$, $c_2 \in ]x,y[$ with the property that $$ \begin{align} f(x + \lambda (y-x)) - f(x) & < \lambda (f(y) - f(x)) \\ \\ \Longrightarrow \quad f'(c_1) = \frac{ f(x + \lambda(y-x)) - f(x)}{\lambda(y-x)} & < \frac{f(y) - f(x)}{y-x} = f'(c_2) \end{align} $$ which is a contradiction. One way to ensure that a function has a strictly decreasing derivative is strict concavity : a function is strictly concave if equality is reversed in the definition of strict convexity and the proof is similar. Is this what you were looking for? I hope it helps.