Monotonicity of maximum of convex combination of two scaled concave functions

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?

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 $$\frac1{A_*(a,b)} and then obviously $$\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.

• Thanks for the counterexample! How about if the maximizer of 𝑓 is positive? Would the inequality then hold? I'm having trouble thinking of a counterexample in such a case... Jan 5 '20 at 10:35
• @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 $\frac{dg^*(b)}{db}=(1/a-2)AB>0$. Jan 5 '20 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, $$\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta) = 0$$.

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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.