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

  • $\begingroup$ Not all of the derivatives of $x^n$ are strictly increasing. Any idea what sort of conditions you're looking for? For what it's worth, if $f^{(n)}$ is increasing, then $f^{(n-1)}$ is convex, so your condition implies that all derivatives of $f$ are convex. $\endgroup$ – Jonas Meyer Dec 5 '11 at 4:03
  • $\begingroup$ There's an obvious pattern the higher derivatives of the example functions possess. For example on the domain $(-\frac{\pi}{2},0)$, the first derivative of $tan(x)$ is decreasing, the 2nd derivative is increasing, the third decreasing and so on. Then on $(0,\frac{\pi}{2})$ they are all increasing. I want to know whats behind this pattern. I know of other functions (which are also increasing ans convex) who's derivatives follow similar patterns. So I thought maybe by knowing something about $f$ one could deduce that all its higher derivatives are increasing! Maybe its not the case tho! $\endgroup$ – mark Dec 5 '11 at 4:33
  • $\begingroup$ In most calculus courses the reverse is deduced, but in most advanced analysis courses it is important to know what can we say about $f'$ if we know things about $f$ : convex functions have important differentiability properties, in particular that they are always semi-differentiable, i.e. differentiable in every direction (the limit when $h \to 0$ is only taken with $h \to 0^+$ though). $\endgroup$ – Patrick Da Silva Dec 5 '11 at 5:03

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.

  • $\begingroup$ Thank you. So strict concavity $<=>$ $f'(x)$ is strictly increasing. But what about the higher derivatives $f^{(n)}(x)$? When are they all increasing or alternatingly increasing and decreasing as in the example above? That's what I was trying to ask. I suppose if one could prove under some condition $f'(x)$ was also convex (as well as increasing as we know) then $f''(x)$ would be increasing and we could inductively apply the "condition" to prove $f^{(n)}(x)$ is increasing. $\endgroup$ – mark Dec 5 '11 at 5:46
  • $\begingroup$ What I said is that when $f$ is differentiable, the strict convexity of $f$ implies that $f'$ is strictly increasing. (You said concavity I think you typed and/or read it wrong.) The converse is also true ; If $f'$ is strictly increasing then $f$ is strictly convex, that is easy enough to prove using the mean value theorem. Why are you so interested in going to more higher powers of the derivative? This seems like a very hard problem to deduce properties of $f^(n)(x)$ by only having constraints on $f$ the way you want it, and unless there's a purpose I don't know where to look. $\endgroup$ – Patrick Da Silva Dec 5 '11 at 7:17
  • $\begingroup$ Yeah, that was a typo. Thanks for your input. As for the motivation see comment above. $\endgroup$ – mark Dec 5 '11 at 17:08
  • $\begingroup$ You can use \Longleftrightarrow to produce this $\Longleftrightarrow$. $\endgroup$ – Patrick Da Silva Dec 5 '11 at 17:12

Let $V$ be the vector space of solutions to a linear differential equation with constant coefficients. Then differentiation is a linear operator $D:V\to V$.

Suppose that $f\in V$ is an eigenvector of $D$ with positive eigenvalue. Then, clearly, if $f$ is increasing and convex, so are $Df, D^2f,\ldots$

This is why $e^{ax}$ has this property; it is the (eigen) solution to the differential equation $Df = af$.

The reason that $x^n$ almost has this property is that it is almost an eigenvector. Specifically, in the space of solutions to the equation $D^{n+1} f = 0$, $x^n$ is a generalized eigenvector of $D$. Specifically, its derivatives give a basis for $V$: $\{x^n,nx^{n-1},\ldots\}$. Note that each entry in this basis is strictly increasing, and convex on $(0,\infty)$. And so positive combinations of them will be also.

The situation with $\tan x$ is a little more complicated, but not much. Both $\tan x$ and $\sec x$ are increasing on $(0,\pi/2)$, and every derivative of $\tan x$ is a polynomial in $\tan x$ and $\sec x$ with positive coefficients. So each time we differentiate, we're remaining in a space of functions which are increasing and convex.

So the answer to this question appears to be: if $f_1,\ldots , f_n$ are increasing, convex functions, and for all $i$, $Df_i$ is a non-zero polynomial in $f_1,\ldots , f_n$ with positive coefficients, then, for any non-zero polynomial $g$ in $f_1,\ldots , f_n$ with positive coefficients, every derivative of $g$ is increasing and convex. In other words, I interpret this as a question as having more to do with algebra than analysis.


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