How to upper bound $F(\alpha)=\max_{z\in[0,1]} z\cdot \min \left\{-\ln(1-e^{-z}),-\alpha\cdot \ln(1-e^{-\alpha z})\right\}$?

Question

Let $\alpha\ge 1$.

I'm looking for an upper bound on the expression $$F(\alpha)=\max_{z\in[0,1]} z\cdot \min \left\{-\ln(1-e^{-z}),-\alpha\cdot \ln(1-e^{-\alpha z})\right\}.$$

---

When $\alpha=1$, we have that $F(1)=\ln^2 2$, using $z=\ln 2$.

For larger $\alpha$ values, $z$ (and $F(\alpha)$) should be lower, but I'm not sure how to get a better bound.

Answer 1

Let $$f(z;\alpha) = -\alpha z \log (1 - e^{-\alpha z}).$$ For $\alpha > 1$, we are motivated to compute the intersection of the curves $f(z;1)$ and $f(z;\alpha)$ for $0 < z < 1$. To this end, we want $$1 = \frac{f(z;\alpha)}{f(z;1)} = \alpha \frac{\log (1 - (e^{-z})^\alpha)}{\log (1 - e^{-z})} $$ which suggests letting $y = e^{-z}$ and attempting to solve for $y > 0$ $$\log (1 - y) = \log (1 - y^\alpha)^\alpha$$ or $$1 - y = (1-y^\alpha)^\alpha.$$ This clearly demonstrates that at best we cannot hope for a closed form solution unless $\alpha \in \mathbb Z^+$, and even then, it appears only the case $\alpha = 2$ admits a closed form. However, it is not intractable to compute $y$ numerically, e.g. via Newton's method. Then we would take $z^* = -\log y$, and then one can easily show that $$F(\alpha) = -z^* \log (1 - e^{-z^*}).$$

As for an upper bound, we want to find some function $U(\alpha)$ for which $F(\alpha) \le U(\alpha)$, which suggests we want to find a $\hat z$ that errs on the side of being larger than $z^*$ but has a closed form. This requires choosing $y$ smaller than the root of $(1-y^\alpha)^\alpha + y - 1$. Heuristics suggest that $y = 1 - (1 + \alpha^{1/2})^{-1}$ works reasonably well, thus $$U(\alpha) = \log \frac{\sqrt{\alpha}}{1 + \sqrt{\alpha}} \log \frac{1}{1 + \sqrt{\alpha}}$$ is a crude upper bound.

We can do more numerics on this upper bound to get something tight, e.g. $$U(\alpha ) = \log \left(-0.0203842 \log ^2 a + 0.194613 \log a + 0.492474\right) \times \\ \log \left(0.0203842 \log ^2 a -0.194613 \log a +0.507526\right)$$ but this is not really based in any theoretic calculations.

Answer 2

Suggestion: Sometimes a good picture is helpful. Below you have the graphs of $f_a(x)=-ax\log(1-e^{-ax})$ for $a=1,2,3,5$. There seems to be a point (to be determined) $x^*_a$ such that $f_a(x)> f_1(x)$ for $0<x<x^*_a$ an $f_a(x)\leq f_1(x)$ for $x^*_a\leq x\leq 1$. If that point can be determined or approximated, then you can compare the maximum of $f_1(x)$ in $[0,x^*_a]$ against the maximum value of $f_2$ in $[x^*_a,1]$.
From the picture, it seems that the value of interest occurs at $x^*_a$.

Plenty of work to do though.

1. Proving the existence of such $x^*_a$ is not difficult since $f_a(0)=0$, $f_a(1)=-a\log(1-e^{-a})\leq -\log(1-e^{-1})$ for $a>1$, $f'_a(0)$ should be also easy to estimate to show by convexity arguments that initially $f_a>f_1$.

2. Compare the location of critical points of $f_a$ and $f_1$.

3. I don't thing close form solution for $x^*_a$ is possible.

4. Careful asymptotics for $x^*_a$ may be possible to obtained.

5. A numerical scheme can be easily implemented to find $x^*_a$ for different values of $a$.

Here is a quick and dirty numerical implementation in R to find the maximum values ($x^*_a$) for $1<a$.

library(latex2exp)
x <- seq(0,1,length.out = 1000)

myfun_cond <- function(x,a){
    ifelse(x==0,0,-a*x*log(1-exp(-a*x)))
}

myfun <- Vectorize(myfun_cond,vectorize.args = c("x","a"))

f <- function(x,a){
    myfun(x,1) - myfun(x,a)
}
fV <- Vectorize(f,vectorize.args = c("x","a"))

xstar_a <- function(a){
    uniroot(fV, c(0.001, 1), a= a)$root
}

xstar_avec <- Vectorize(xstar_a,vectorize.args = "a")

avalues <- seq(1.02,15,length.out = 200)

plot(avalues,xstar_avec(avalues),type = 'l',
         xlab = "a", ylab = TeX("$x^*_a$"),
         main = TeX("maximum points"))

Here is a plot between the numerically estimate maximum value for $a>1$ and the raw upper bound found by heroup