# Computing minimum / maximum of strange two variable funcion

I want to compute $$\max_{\frac{1}{5 \theta }\leq \alpha \leq \frac{1}{2}} \left(\frac{\alpha\log (\alpha)}{1-\alpha} + \log \left( 1 - \alpha\right) + \frac{1}{1-\alpha} \cdot \left( f\left(\frac{1-\alpha}{2\alpha}\right) + 1 + \frac{41}{15\theta} \right)\right)$$ for \begin{align*} f(x) = \begin{cases} - 1.3- \log x + x/2 +1/2 - 1/x&\text{for } x \geq 2\\ -1.3 &\text{otherwise}. \end{cases} \end{align*}

Here, $\log$ is base 2. Intuitively, it should be either for $\alpha=1/2$ or for $\alpha = 1/(5\theta)$ (and a plot verifies this). Is there a way how to prove this?

In the end I want to compute the $\theta \geq 1$ where the whole expression reaches its minimum. Could there be another approach to do this?

• What values may $\theta$ take? – Iosif Pinelis Jul 24 '18 at 18:12

Let $a:=\alpha$ and $u:=\frac1{5\theta}$. The condition $\theta\ge1$ (now added in the question) means that $0<u\le1/5$, which will be assumed henceforth. We need to compute $$\inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a),$$ where \begin{equation*} F(u,a):=\begin{cases} F_1(u,a)&\text{ if }1/5\le a\le1/2,\\ F_2(u,a)&\text{ if }u\le a\le1/5, \end{cases} \end{equation*} $$F_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2},$$ $$F_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2},$$ \begin{multline} H(u,a):=-60 a^3 \ln a-820 a^2 u \ln2+60 a^2 \ln a-117 a^2 \ln2+820 a u \ln2 \\ +60 (a-1) a \ln \left(\frac{1}{2} \left(\frac{1}{a}-1\right)\right)+60 (a-1)^2 a \ln (1-a)-18 a \ln2+15 \ln2. \end{multline} So, the infsup in question is $$\inf_{0<u\le1/5}(M_1(u)\vee M_2(u)),$$ $$M_1(u):=\sup_{1/5\le a\le1/2}F_1(u,a),\quad M_2(u):=\sup_{a\in[u,1/5]}F_2(u,a).$$

For $F_1(a):=F_1(u,a)$, let $DF_1(a):=F_1'(a)(1-a)^2$. Then $DF_1(a)=-3/10 + 41 u/3 + \ln a/\ln2$ is increasing in $a$. So, $DF_1(a)$ (and hence $F_1'(a)$) can change the sign only from $-$ to $+$. So, \begin{align} M_1(u)&=\sup_{1/5\le a\le1/2}F_1(u,a)=F_1(u,1/5)\vee F_1(u,1/2). \end{align}

For $F_2(a):=F_2(u,a)$, let $DF_2(a):=F_2'(a)(1-a)^2$. Then $DF_2'(a) 2 \ln2\,(1-a)^2 a^3=2 a^4 + \ln2 - a (2 + \ln8) + a^2 (8 + \ln8) - a^3 (8 + \ln512)>0$ for $a\in[0,1/5]$. So, $DF_2(a)$ is increasing in $a\in[0,1/5]$, and so, $DF_2(a)$ (and hence $F_2'(a)$) can change the sign only from $-$ to $+$. So, $$M_2(u)=\sup_{a\in[u,1/5]}F_2(u,a)= F_2(u,u)\vee F_2(u,1/5)\quad\text{if }0<u\le1/5.$$

Therefore and because $F_1(u,1/5)=F_2(u,1/5)$, the infsup in question is $$\inf_{0<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)].$$

For brevity, let $$g(u):=F_2(u,u),\quad d_a(u):=F_2(u,u)-F_1(u,a)=g(u)-F_1(u,a).$$ Let $g_1(u):=g'(u)(1 - u)^2$. Then $g_1'(u)$ is a simple rational function of $u$, which is $>0$ (everywhere here $0<u\le1/5$). So, $g_1(u)$ is increasing in $u$. So, $g_1(u)$ (and hence $g'(u)$) can change the sign only from $-$ to $+$. Now we find $$\inf_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=1.0616\ldots,$$ attained at $u=0.099677\ldots$.

Because (i) $d_a(u)=g(u)-F_1(u,a)$, (ii) $F_1(u,a)$ is affine in $u$, and (iii) $g'(u)$ can change the sign only from $-$ to $+$, we see that $d_a'(u)$ can change the sign only from $-$ to $+$. Also, $d_{1/5}'(14/100)=-6.7650\ldots<0$. So, $d_{1/5}$ is decreasing on $[0,14/100]$, with $d_{1/5}(14/100)=0.1884\ldots>0$. So, $d_{1/5}>0$ on $[0,14/100]$, that is, $F_2(u,u)>F_1(u,1/5)$ if $0<u\le14/100$. Similarly, using that $d_{1/2}'(14/100)=-17.015\ldots<0$ and $d_{1/2}(14/100)=0.07602\ldots>0$, we verify that $F_2(u,u)>F_1(u,1/2)$ if $0<u\le14/100$.

So, $$\inf_{0<u\le14/100}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]= \min_{0<u\le14/100}F_2(u,u)= 1.0616\ldots,$$ attained at $u=0.099677\ldots$. On the other hand, because $F_1(u,1/2)$ is increasing in $u$, we have \begin{multline} \inf_{14/100<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]\ge\inf_{14/100<u\le1/5}F_1(u,1/2) \\ = F_1(14/100,1/2) =1.2266\ldots \\ >1.0616\ldots =\inf_{0<u\le14/100}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]. \end{multline}

Thus, with $M(u):=F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)$, the infsup in question is \begin{multline} \inf_{0<u\le1/5}M(u) =\inf_{0<u\le14/100}M(u) \bigwedge \inf_{14/100<u\le1/5}M(u)\\ =\inf_{0<u\le14/100}M(u)= 1.0616\ldots, \end{multline} attained at $a=u=0.099677\ldots$.

• Thanks, a lot!! Sorry, I was not precise: $\theta \geq 1$ (so $u\geq 1/5$) and the logarithm is base 2. – Armin Weiß Jul 25 '18 at 11:45
• I am still not quite sure: do you want $\theta\ge1$ or $\theta\le1$? Because $\theta\ge1$ corresponds to $u\le1/5$, not $u\ge1/5$. In any case, I think your specification the values of $\theta$ of interest would only simplify the proof. Also, in mathematical literature, $\log$ without a specification of the base usually means the natural log (I think $\ln$ is better for that purpose). – Iosif Pinelis Jul 25 '18 at 15:25
• oops... $\theta \geq 1$ (so $u\leq 1/5$) – Armin Weiß Jul 25 '18 at 15:35
• I have modified the answer, to take into account that $\theta\ge1$ and the log is base $2$. – Iosif Pinelis Jul 25 '18 at 17:14
• thanks for the change! ...In my calculation $F(u,a) = 1.06589...$ for $a=u=0.095260$ though ($a=u=0.095260$ seems to be correct)... – Armin Weiß Jul 26 '18 at 14:44