Maximize Complicated Function.

Question

Let $f : (0, +\infty) \to \mathbb{R}$ be the following function

\begin{equation} f(a) = \frac{(2 p^a - (p+\delta)^a - (p - \delta)^a)^2}{4\ p^a (1-p^a)} \end{equation} where $0<p<1,\ 0< p+ \delta < 1,\ 0 < p- \delta< 1$.

Is it possible to obtain a bound for the maximum value of $f$ as a function of $p, \delta$ or determine that $f$ is unbounded (going to $+\infty$) ?

I tried differentiating etc. but the expression of the derivative is too complicated (even for Mathematica) to find its roots.

EDIT:

Using Mlazhinka Shung Gronzalez LeWy's answer we can assume that $p < (p+\delta)^2$ and thus $f$ is bounded. Is it possible to calculate a bounding value for $f$?

Are there any inequalities to upper bound the numerator $(2 p^a - (p+\delta)^a - (p - \delta)^a)^2$ ?

Answer 1

Set $p^a=t$ and the expression becomes ($4$ dropped)

$$\frac{(2t-t^\alpha-t^\beta)^2}{t(1-t)}$$ where $\alpha:=\dfrac{\log(p+\delta)}{\log p}<1$, $\beta:=\dfrac{\log(p-\delta)}{\log p}>1$ and $t\in(0,1)$.

Close to $t=0$, it simplifies to

$$\left(2t^{1/2}-t^{\alpha-1/2}-t^{\beta-1/2}\right)^2,$$ which is unbounded when $\alpha<1/2$.

Again for small $t$, the expression that is squared can be approximated by dropping the term with the highest exponent, giving

$$2t^{1/2}-t^{\alpha-1/2}$$ or $$2u-u^{2\alpha-1}.$$

The latter has an extremum at

$$u=\left(\alpha-\frac12\right)^{-1/(2\alpha-2)},$$

giving an estimate of the maximum

$$\left(\left(\alpha-\frac12\right)^{-1/(\alpha-1)}-\left(\alpha-\frac12\right)^{-(2\alpha-1)/(2\alpha-2)}\right)^2.$$

Below the function under parenthesis and its approximation (in blue):

Answer 2

The boundedness reduces to studying the limits when $a\to0+$ and $a\to\infty$. Let's look at the latter for starters.

Let $a=2n$ with $n$ positive integer. Then

\begin{align} f(2n)&=\frac{(2 p^n - \frac{(p+\delta)^{2n} + (p - \delta)^{2n}}{p^n})^2}{4(1-p^a)} \end{align}

We can assume $\delta\geq0$.

The denominator tends to $4$.

The $2p^n$ tends to zero.

So, the finiteness of the limit of $f(2n)$ is that of the limit of $$\frac{(p+\delta)^{2n}+(p-\delta)^{2n}}{p^n}=\frac{1+\left(\frac{p-\delta}{p+\delta}\right)^{2n}}{\left(\frac{p}{(p+\delta)^2}\right)^n}$$.

the numerator of this tends to $1$. The denominator ... well it depends if $p/(p+\delta)^2$ is greater than or equal, or smaller than 1.