Let $f:R\rightarrow R$ be a concave function with a unique and finite maximum. Let $$g(x, \beta) = \beta f(\alpha \cdot x) + (1-\beta) f((1-\alpha) \cdot x), $$ where $\alpha \in [0,1/2]$ and $\beta \in [0,1/2]$. Furthermore, let $g^*(\beta) = \max_{x} g(x,\beta)$.

I am trying to find conditions for $f$ such that $$ \frac{d}{d\beta} g^*(\beta) \leq 0 $$ for $\alpha \in [0,1/2]$ and $\beta \in [0,1/2]$. Will the inequality above be satisfied if $f$ is concave with a unique and finite maximum? What conditions do I need? Does the inequality hold when the maximizer of $f$, i.e. $\arg \max_x f(x)$, is positive?


2 Answers 2


Let $a:=\alpha\in(0,1/2)$ and $b:=\beta\in(0,1/2)$. Take any positive real $A$ and any real $B$, and let $$f(x):=\min[x,(1+A)B-Ax] $$ for real $x$. Then the function $f$ is concave with a unique and finite maximum (at $x=B$), and $$g^*(b)=g\Big(\frac B{1-a},b\Big)= \Big(1-\frac{1-2 a}{1-a}\,b\Big) B $$ if $a$ and $b$ are small enough so that $$\frac1{A_*(a,b)}<A<A_*(a,b):=\frac{(1-a)(1-b)}{ab}, $$ and then obviously $$\frac{dg^*(b)}{db}=-\frac{1-2 a}{1-a}\, B>0 $$ if $B<0$. So, your desired inequality does not always hold for concave functions $f$ with a unique and finite maximum.

Since, in the above example, the negative slope $-A$ can be any negative real number, it seems unlikely that there exists a simple and good enough sufficient condition for your desired inequality to hold.

  • $\begingroup$ Thanks for the counterexample! How about if the maximizer of đť‘“ is positive? Would the inequality then hold? I'm having trouble thinking of a counterexample in such a case... $\endgroup$
    – ACopt
    Jan 5, 2020 at 10:35
  • $\begingroup$ @ACopt : In the same example, if $B>0$, $0<A<1$, and $a$ and $b$ are close enough to $1/2$ so that $A<1/A_*(a,b)$, then $g^*(b)=g(\frac Ba,b)$ and $\frac{dg^*(b)}{db}=(1/a-2)AB>0$. $\endgroup$ Jan 5, 2020 at 13:51

Remark 1: If $f: \ \mathbb{R} \to \mathbb{R}$ is a twice differentiable function with $f'' < 0$ and finite maximum $f(x_0)$ at $x_0 > 0$. We may analyze the condition by using Danskin's theorem. See https://en.wikipedia.org/wiki/Danskin%27s_theorem

Remark 2: If $\alpha = 0$, we have $g(x, \beta) = \beta f(0) + (1-\beta)f(x)$ and $g^{\ast}(\beta) = \beta f(0) + (1-\beta) f(x_0)$. Thus, $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta) = f(0) - f(x_0) \le 0$.

Remark 3: If $\alpha = \frac{1}{2}$, we have $g(x, \beta) = f(\frac{x}{2})$ and $g^{\ast}(\beta) = f(x_0)$. Thus, $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta) = 0$.


According to Remarks 2 and 3, we may restrict to $\alpha \in (0, \frac{1}{2})$.

Clearly, for any $\beta \in [0, \frac{1}{2}]$, $g(x, \beta)$ has a unique maximizer denoted by $x^\ast(\beta)$, which is the unique solution of $$\alpha\beta f'(\alpha x) + (1-\alpha)(1-\beta)f'((1-\alpha)x) = 0.$$ It implies that $\frac{x_0}{1-\alpha} \le x^\ast(\beta) \le \frac{x_0}{\alpha}$. Thus, we have $g^{\ast}(\beta) = \max_{x\in S} g(x, \beta)$ for some compact set $S$ containing $[\frac{x_0}{1-\alpha}, \frac{x_0}{\alpha}]$, which satisfies the requirement of Danskin's theorem.

By using Danskin's theorem, we have $$\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta) = f(\alpha x^\ast(\beta)) - f((1-\alpha) x^\ast(\beta)).$$

A sufficient and necessary condition for $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta)\le 0, \ \forall \beta \in [0, \frac{1}{2}]$ given $\alpha \in (0, \frac{1}{2})$ is that \begin{align} &\alpha\beta f'(\alpha x) = - (1-\alpha)(1-\beta)f'((1-\alpha)x)\\ \Longrightarrow\quad & f(\alpha x) \le f((1-\alpha) x), \quad \forall \alpha \in (0, \tfrac{1}{2}), \beta \in [0, \tfrac{1}{2}]. \end{align} Maybe we can obtain some sufficient conditions.


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