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$f \in C^2([0,1])$ with $f''$ convex and $f(0) = f'(0) = f''(0) = 0$.

Is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?


Source: AoPS

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    $\begingroup$ What makes you suspect that it might be true? $\endgroup$ Oct 30 '21 at 12:58
  • $\begingroup$ artofproblemsolving.com/community/c7h1570936 $\endgroup$
    – Dattier
    Oct 30 '21 at 13:08
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    $\begingroup$ Any such function $f$ is a third-order-convex function, that is, all of the divided differences $[a,b,c,d,e;f]$ are at least $0$. Compute the limiting divided difference $[0,0,1,1,1;f] = \lim_{\epsilon \rightarrow 0} [0,\epsilon,1-2\epsilon,1-\epsilon,1;f]$ and the inequality pops out. You don't need to assume that $f''(0) = 0$. $\endgroup$
    – zeb
    Oct 30 '21 at 17:28
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The answer is yes, and it follows immediately from a simple description of the extreme rays of the convex cone of the functions with a convex second derivative.

Indeed, for $x\in(0,1]$, let $f'''(x)$ denote the left derivative of the convex function $f''$ at $x$, so that $f'''$ is non-decreasing on $(0,1]$.

Take any $x\in(0,1]$. Then, taking into account the conditions $f(0)=f'(0)=0$, one has the Taylor-like formula (cf. e.g. formula (4.12) or formula (3.2)): \begin{equation*} f(x)=\frac{f''(0)}2\,x^2+\frac{f'''(0+)}6\,x^3+\frac16\int_{(0,1)}df'''(t)(x-t)_+^3, \tag{1} \end{equation*} where $z_+:=\max(0,z)$ and the integral is a Lebesgue--Stieltjes one (see a detailed derivation of this formula at the end of this answer.)

So, $f$ is a "mixture" of the functions $x\mapsto g_2(x):=x^2$, $x\mapsto g_3(x):=x^3$, and $x\mapsto f_t(x):=(x-t)_+^3$ for $t\in(0,1)$, and the "coefficients" of the functions $f_t$ in this mixture are nonnegative.

Therefore, it remains to note that (i) $f''(1)+6f(1)=4f'(1)$ if $f=g_2$ or $f=g_3$ and (ii) $f''(1)+6f(1)\geq 4f'(1)$ if $f=f_t$ for some $t\in(0,1)$ (in the latter case, the inequality $f''(1)+6f(1)\ge 4f'(1)$ is the obvious inequality $6z+6z^3\ge12z^2$ with $z:=1-t$). $\qquad\Box$

The condition $f''(0)=0$ was not needed or used here.


Details on the Taylor-like formula (1):

Take any $x\in(0,1]$. Then \begin{equation*} \begin{aligned} f''(x)&=f''(0)+\int_0^x du\,f'''(u) \\ &=f''(0)+\int_0^x du\,\Big(f'''(0+)+\int_{(0,u)}df'''(t)\Big) \\ &=f''(0)+f'''(0+)x+\int_0^x du\,\int_{(0,u)}df'''(t) \\ &=f''(0)+f'''(0+)x+\int_{(0,x)}df'''(t)\int_t^x du\, \\ &=f''(0)+f'''(0+)x+\int_{(0,x)}df'''(t)(x-t) \\ &=f''(0)+f'''(0+)x+\int_{(0,1)}df'''(t)(x-t)_+; \end{aligned} \end{equation*} the fourth equality in the above display is an instance of the Fubini--Tonelli theorem.

Therefore and by a Taylor formula and the conditions $f(0)=f'(0)=0$, \begin{equation*} \begin{aligned} f(x)&=\int_0^x du\,f''(u)(x-u) \\ &=\int_0^x du\,\Big(f''(0)+f'''(0+)x+\int_{(0,1)}df'''(t)(x-t)_+\Big)(x-u) \\ &= \frac{f''(0)}2\,x^2+\frac{f'''(0+)}6\,x^3+\frac16\int_{(0,1)}df'''(t)(x-t)_+^3. \end{aligned} \end{equation*} One can also check this Taylor-like formula by taking the integral $\int_{(0,1)}df'''(t)(x-t)_+^3$ "by parts" three times -- that is, by applying the Fubini--Tonelli theorem to this integral three times.

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The inequality you wrote down is a special case of a general principle about higher-order convexity: any "simple enough" linear inequality of the type you wrote down will be true as long as it is true for every polynomial function which satisfies your convexity condition. Furthermore, these simple inequalities can always be proved directly by a divided difference computation.

To make this claim precise, I have to define higher-order convexity, and clarify what I mean by "simple enough". I'll start by reviewing the elementary theory of divided differences. Divided differences are defined inductively by the rules

  • $[a;f] = f(a)$,
  • $[a,b;f] = \frac{f(a) - f(b)}{a-b}$,
  • $[a_0, ..., a_n, a_{n+1}; f] = \frac{[a_0, ..., a_{n-1}, a_n; f] - [a_0, ..., a_{n-1}, a_{n+1}; f]}{a_n - a_{n+1}}$ if $a_n \ne a_{n+1}$.

The divided difference $[a_0, \dotsc, a_n; f]$ turns out to be a symmetric function of the points $a_0, \dotsc, a_n$ — in fact, we have the explicit formula $$ [a_0, \dotsc, a_n; f] = \sum_{i=0}^n \frac{f(a_i)}{\prod_{j\ne i}(a_i - a_j)}, $$ which can easily be proved by induction. If $p(x)$ is a polynomial of degree $n$ with leading coefficient $cx^n$, then the divided difference $[a_0, \dotsc, a_n; p]$ will be equal to the constant $c$.

A function $f$ is called $(k-1)$th-order convex if every divided difference $[a_0, \dotsc, a_k; f]$ is positive, so ordinary convexity corresponds to first-order convexity. For $k \ge 3$, something nice happens:

Proposition: If $k \ge 3$, then $f$ is $(k-1)$th-order convex if and only if $f$ is differentiable and $f'$ is $(k-2)$th-order convex.

This fact can be proved directly — it's a fun exercise. As a consequence, a function $f$ is $(k-1)$th-order convex for $k \ge 2$ if and only if $f \in C^{k-2}$ and $f^{(k-2)}$ is convex. So the functions that showed up in this question in are exactly the third-order convex functions on $[0,1]$.

Now for the sorts of inequalities we'd like to consider: suppose that $a_i$, $b_i$, $c_i$ are constants with $b_i \in \mathbb{N}$, and consider the functional inequality $$ \sum_i c_if^{(b_i)}(a_i) \stackrel{?}{\ge} 0.\tag{$*$}\label{star} $$ Define the complexity of the inequality \eqref{star} to be $$ \mathcal{C}(*) = \sum_{a \in \{a_i\}} \max\{b_i+1 \mid a_i = a\}. $$ The general fact is:

Theorem: If $\mathcal{C}(*) \le k+1$, then the following are equivalent:

  • the inequality \eqref{star} is true for all $(k-1)$th-order convex functions $f$,
  • the inequality \eqref{star} is an equality when we plug in $f(x) = x^m$ for $m < k$, and \eqref{star} is true when we plug in $f(x) = x^k$,
  • there is a constant $C \ge 0$ such that $\sum_i c_if^{(b_i)}(a_i) = C\cdot[\underbrace{a, \dotsc, a,}_{\text{$\max\{b_i+1 \mid a_i = a\}$ times}} \dotsc; f]$, where the divided difference is interpreted as a limit of divided differences at infinitesimally shifted points if any $b_i > 0$.

The proof of this general fact ends up being a straightforward exercise in linear algebra.

Things get a bit more complex once we jump to $\mathcal{C}(*) \le k+2$ — I used to be very interested in these sorts of inequalities a long time ago, and I have an article Inequalities and higher order convexity on the arXiv with lots of tricks for verifying these sorts of low complexity functional inequalities.

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    $\begingroup$ I am wondering if your theorem can be deduced from the Farkas lemma (en.wikipedia.org/wiki/Farkas%27_lemma) and/or using the generalized Vandermonde matrix (jstor.org/stable/2690290). Also, wondering if $f(x)=x^k$ in your theorem can be replaced by $f(x)=e^x$ or, more generally, by $f(x)=g(x)$, where $g$ is any given $(k−1)$th-order strictly convex function. $\endgroup$ Oct 31 '21 at 11:57
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    $\begingroup$ @Iosif The function $f(x) = x^k$ can indeed be replaced by any given $(k-1)$th-order strictly convex function - the only purpose of it is to make sure that you are getting the direction of the inequality right! Farkas lemma is unnecessary here, all you need to do is to check that the system of linear equations (satisfied by the $c_i$) which you get by checking that you have equality at $f(x) = x^m$ for $m < k$ is non-degenerate, so that it has just one solution, up to scale. $\endgroup$
    – zeb
    Oct 31 '21 at 14:25
  • $\begingroup$ Thanks for the valuable insight about convexity and divided differences - this is one of the best answers I came across on Mathoverflow. $\endgroup$
    – pinaki
    Nov 1 '21 at 14:05
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$\newcommand\R{\mathbb R}$The purpose of this answer is to provide a short, simple, and self-contained proof of the theorem in zeb's answer.

For (say) a nonempty open interval $I\subseteq\R$ and sufficiently smooth functions $f\colon I\to\R$, let \begin{equation*} Lf:=\sum_{i\in[m]}\sum_{j\in[n_i-1]_0}c_{i,j}f^{(j)}(a_i), \end{equation*} where $m$ and the $n_i$'s are natural numbers, $[m]:=\{1,\dots,m\}$, $[m]_0:=\{0,\dots,m\}$, the $c_{i,j}$'s are real numbers such that $c_{i,n_i-1}\ne0$, $f^{(j)}$ is the $j$th derivative of $f$, and the $a_i$'s are distinct points in $I$. Let also \begin{equation*} k:=-1+\sum_{i\in[m]}n_i, \tag{0} \end{equation*} and consider the divided difference \begin{equation*} L_0f:=[a_1^{,n_1},\dots,a_m^{,n_m};f], \end{equation*} where $a^{,n}:=\underbrace{a,\dots,a}_{n}$. Finally, let $C_{k-1}^+$ denote the set of all $(k-1)$th-order convex functions and let $p_r(x):=x^r$.

Then the mentioned theorem can be restated as follows:

Theorem 1: The following three statements are equivalent to one another:

(i) $Lf\ge0$ for all $f\in C_{k-1}^+$.

(ii) $Lp_r=0$ for all $r\in[k-1]_0$ and $Lp_k\ge0$.

(iii) $L=cL_0$ for some real $c\ge0$.

Proof: Implication (i)$\implies$(ii) is trivial.

By approximation, we may assume that the functions $f$ of interest are however smooth. So, in the case when $n_i=1$ for all $i$, implication (iii)$\implies$(i) follows by induction, using the mean value theorem. If $n_i>1$ for some or all $i$, implication (iii)$\implies$(i) follows then by a limit transition.

It remains to check implication (ii)$\implies$(iii). The part ($Lp_r=0$ for all $r\in[k-1]_0$) of statement (ii) can be rewritten as the system of equations \begin{equation*} \sum_{(i,j)\in K}a_{r,(i,j)}c_{i,j}=-c_{m,n_m-1}p_r^{(n_m-1)}(a_m)\text{ for }r\in[k-1]_0 \tag{1} \end{equation*} with the "unknowns" $c_{i,j}$ for $(i,j)\in K$, where \begin{equation*} K:=\{(i,j)\colon i\in[m]\ \&\ j\in[n_i-1]_0\}\setminus\{(m,n_m-1)\} \end{equation*} and $a_{r,(i,j)}:=p_r^{(j)}(a_i)$. The matrix $A:=(a_{r,(i,j)}\colon r\in[k-1]_0,(i,j)\in K)$ of system (1) is square (by (0)) and nonsingular -- otherwise, some nontrivial linear combination of the rows of $A$ is the zero row; that is, for some real $b_r$'s not all of which are $0$, the polynomial $P:=\sum_{r\in[k-1]_0}b_rp_r$ of degree $\le k-1$ would have $k$ roots (namely, $a_1^{,n_1},\dots,a_m^{,n_m-1}$, counting the multiplicities).

So, for each given value of $c_{m,n_m-1}$, system (1) has a unique solution $(c_{i,j}\colon(i,j)\in K)$. On the other hand, $L_0p_r=0$ for all $r\in[k-1]_0$. So, $L=cL_0$ for some real $c$. Moreover, $L_0p_k>0$ and hence the condition $Lp_k\ge0$ in (ii) yields $c\ge0$. $\quad\Box$.

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Here is a more problem solving approach, since this problem comes from AoPS. Instead of working with $f$, work with $g = f''$ which is a convex function.

  1. Note that it is enough to assume $g(1)=0$. Indeed, take $h(x)=f(x)-\frac{x^2}{2} f''(1)$. Then $h'(x) = f'(x)-xf''(1), h''(x) = f''(x)-f''(1)$. Therefore $h''$ is convex is equivalent to $f''$ convex. Moreover $$ h''(1)+6h(1)-4h'(1) = f''(1)+6f(1)-4f'(1)$$ so the desired inequality for $h$ is equivalent with the one for $f$. Furthermore $h''(1)=0$.

  2. Assuming that $g(1)=0$ and $g$ is convex, we see that $f''=g, f(0)=f'(0)=0$ implies $$ f'(t) = \int_0^t g(s)ds, f(t) =\int_0^t \int_0^s g(\tau) d\tau$$ Therefore, it is enough to prove that $$ \int_0^1 \int_0^t g(s)ds \geq \frac{2}{3} \int_0^1 g(t)dt.$$ Denoting $h(t) = \int_0^t g(s)ds$ we have $h'(0)=h'(1)=0$ and $h'\leq 0$ since $g$ is convex on $[0,1]$ with $g(0)=g(1)=0$. Then of course $$ \int_0^1 h(t)dt \geq 2/3 h(1)$$ since $h(1)$ is the minimal value of $h$ on $[0,1]$.

Remark: Compared to the previous answers, I used $f''(0)=g(0)=0$. However, I think it is possible to make a trick similar to point 1 above to see that a similar result holds in the general case.

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