Return to Answer

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 \begin{equation} \inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a), \end{equation} 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*} \begin{equation} F_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2}, \end{equation} \begin{equation} F_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2}, \end{equation} \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 \begin{equation} \inf_{0<u\le1/5}(M_1(u)\vee M_2(u)), \end{equation} \begin{equation} 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). \end{equation}

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, \begin{equation} 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. \end{equation}

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

For brevity, let \begin{equation} g(u):=F_2(u,u),\quad d_a(u):=F_2(u,u)-F_1(u,a)=g(u)-F_1(u,a). \end{equation} 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 \begin{equation} \inf_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=1.0616\ldots, \end{equation} 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, \begin{equation} \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, \end{equation} 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_{13/100<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, the infsup in question is \begin{multline} \inf_{0<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)] \\ =\inf_{0<u\le14/100}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)] \\ \bigwedge \inf_{14/100<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]\\ =\inf_{0<u\le14/100}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]= 1.0616\ldots, \end{multline} attained at $a=u=0.099677\ldots$.

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 \begin{equation} \inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a), \end{equation} 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*} \begin{equation} F_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2}, \end{equation} \begin{equation} F_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2}, \end{equation} \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 \begin{equation} \inf_{0<u\le1/5}(M_1(u)\vee M_2(u)), \end{equation} \begin{equation} 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). \end{equation}

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, \begin{equation} 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. \end{equation}

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

For brevity, let \begin{equation} g(u):=F_2(u,u),\quad d_a(u):=F_2(u,u)-F_1(u,a)=g(u)-F_1(u,a). \end{equation} 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 \begin{equation} \inf_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=1.0616\ldots, \end{equation} 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, \begin{equation} \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, \end{equation} 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 $a=u=0.099677\ldots$.

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 \begin{equation} \inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a), \end{equation} where \begin{equation*} F(u,a):=\begin{cases} F_1(u,a)&\text{ if }1/5<a\le1/2,\\ F_2(u,a)&\text{ if }u\le a\le1/5, \end{cases} \end{equation*} \begin{equation} F_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2}, \end{equation} \begin{equation} F_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2}, \end{equation} \begin{multline} H(u,a):=-60 a^3 \ln a-820 a^2 u \ln2+60 a^2 \ln a-213 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)+78 a \ln2+15 \ln2. \end{multline} So, the infsup in question is \begin{equation} \inf_{0<u\le1/5}(M_1(u)\vee M_2(u)), \end{equation} \begin{equation} M_1(u):=\sup_{1/5<a\le1/2}F_1(u,a),\quad M_2(u):=\sup_{a\in[u,1/5]}F_2(u,a). \end{equation}

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<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, \begin{equation} 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. \end{equation}

So, the infsup in question is \begin{equation} \inf_{0<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)\vee F_2(u,1/5)]. \end{equation} Since $F_1(u,1/5), F_1(u,1/2), F_2(u,1/5)$ are affine in $u$, it is easy to see that $F_1(u,1/5)\vee F_1(u,1/2)< F_2(u,1/5)$ if $0<u\le1/5$. So,
the infsup in question is \begin{equation} \inf_{0<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]. \end{equation} For brevity, let \begin{equation} g(u):=F_2(u,u),\quad d(u):=F_2(u,u)-F_2(u,1/5)=g(u)-F_2(u,1/5). \end{equation} 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 \begin{equation} \inf_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=2.8343\ldots, \end{equation} attained at $u=0.095260\ldots$.

Because (i) $d(u)=g(u)-F_2(u,1/5)$, (ii) $F_2(u,1/5)$ is affine in $u$, and (iii) $g'(u)$ can change the sign only from $-$ to $+$, we see that $d'(u)$ can change the sign only from $-$ to $+$. Also, $d'(13/100)=-6.4029\ldots<0$. So, $d$ is decreasing on $[0,13/100]$, with $d(13/100)=0.10364\ldots>0$. So, $d>0$ on $[0,13/100]$, that is, $F_2(u,u)>F_2(u,1/5)$ if $0<u\le13/100$. Thus, \begin{equation} \inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]= \min_{0<u\le13/100}F_2(u,u)= 2.8343\ldots, \end{equation} attained at $u=0.095260\ldots$. On the other hand, because $F_2(u,1/5)]$ is increasing in $u$, we have \begin{multline} \inf_{13/100<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]\ge\inf_{13/100<u\le1/5}F_2(u,1/5) \\ = F_2(13/100,1/5) =2.9434\ldots>2.8343\ldots =\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]. \end{multline}

Thus, the infsup in question is \begin{multline} \inf_{0<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)] \\ =\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]\bigwedge \inf_{13/100<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]\\ =\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]= 2.8343\ldots, \end{multline} attained at $a=u=0.095260\ldots$.

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 \begin{equation} \inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a), \end{equation} 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*} \begin{equation} F_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2}, \end{equation} \begin{equation} F_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2}, \end{equation} \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 \begin{equation} \inf_{0<u\le1/5}(M_1(u)\vee M_2(u)), \end{equation} \begin{equation} 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). \end{equation}

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, \begin{equation} 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. \end{equation}

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

For brevity, let \begin{equation} g(u):=F_2(u,u),\quad d_a(u):=F_2(u,u)-F_1(u,a)=g(u)-F_1(u,a). \end{equation} 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 \begin{equation} \inf_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=1.0616\ldots, \end{equation} 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, \begin{equation} \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, \end{equation} 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_{13/100<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, the infsup in question is \begin{multline} \inf_{0<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)] \\ =\inf_{0<u\le14/100}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)] \\ \bigwedge \inf_{14/100<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]\\ =\inf_{0<u\le14/100}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]= 1.0616\ldots, \end{multline} attained at $a=u=0.099677\ldots$.

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 \begin{equation} \inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a), \end{equation} where \begin{equation*} F(u,a):=\begin{cases} F_1(u,a)&\text{ if }1/5<a\le1/2,\\ F_2(u,a)&\text{ if }u\le a\le1/5, \end{cases} \end{equation*} \begin{equation} F_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2}, \end{equation} \begin{equation} F_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2}, \end{equation} \begin{multline} H(u,a):=-60 a^3 \ln a-820 a^2 u \ln2+60 a^2 \ln a-213 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)+78 a \ln2+15 \ln2. \end{multline} So, the infsup in question is \begin{equation} \inf_{0<u\le1/5}(M_1(u)\vee M_2(u)), \end{equation} \begin{equation} M_1(u):=\sup_{1/5<a\le1/2}F_1(u,a),\quad M_2(u):=\sup_{a\in[u,1/5]}F_2(u,a). \end{equation}

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<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, \begin{equation} 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. \end{equation}

So, the infsup in question is \begin{equation} \inf_{0<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)\vee F_2(u,1/5)]. \end{equation} Since $F_1(u,1/5), F_1(u,1/2), F_2(u,1/5)$ are affine in $u$, it is easy to see that $F_1(u,1/5)\vee F_1(u,1/2)< F_2(u,1/5)$ if $0<u\le1/5$. So,
the infsup in question is \begin{equation} \inf_{0<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]. \end{equation} For brevity, let \begin{equation} g(u):=F_2(u,u),\quad d(u):=F_2(u,u)-F_2(u,1/5)=g(u)-F_2(u,1/5). \end{equation} 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 \begin{equation} \inf_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=2.8343\ldots, \end{equation} attained at $u=0.095260\ldots$.

Because (i) $d(u)=g(u)-F_2(u,1/5)$, (ii) $F_2(u,1/5)$ is affine in $u$, and (iii) $g'(u)$ can change the sign only from $-$ to $+$, we see that $d'(u)$ can change the sign only from $-$ to $+$. Also, $d'(13/100)=-6.4029\ldots<0$. So, $d$ is decreasing on $[0,13/100]$, with $d(13/100)=0.10364\ldots>0$. So, $d>0$ on $[0,13/100]$, that is, $F_2(u,u)>F_2(u,1/5)$ if $0<u\le13/100$. Thus, \begin{equation} \inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]= \min_{0<u\le13/100}F_2(u,u)= 2.8343\ldots, \end{equation} attained at $u=0.095260\ldots$. On the other hand, because $F_2(u,1/5)]$ is increasing in $u$, we have \begin{multline} \inf_{13/100<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]\ge\inf_{13/100<u\le1/5}F_2(u,1/5) \\ = F_2(13/100,1/5) =2.9434\ldots>2.8343\ldots =\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]. \end{multline}

Thus, the infsup in question is \begin{multline} \inf_{0<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)] \\ =\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]\bigwedge \inf_{13/100<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]\\ = 2.8343\ldots, \end{multline} attained at $a=u=0.095260\ldots$.

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 \begin{equation} \inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a), \end{equation} where \begin{equation*} F(u,a):=\begin{cases} F_1(u,a)&\text{ if }1/5<a\le1/2,\\ F_2(u,a)&\text{ if }u\le a\le1/5, \end{cases} \end{equation*} \begin{equation} F_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2}, \end{equation} \begin{equation} F_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2}, \end{equation} \begin{multline} H(u,a):=-60 a^3 \ln a-820 a^2 u \ln2+60 a^2 \ln a-213 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)+78 a \ln2+15 \ln2. \end{multline} So, the infsup in question is \begin{equation} \inf_{0<u\le1/5}(M_1(u)\vee M_2(u)), \end{equation} \begin{equation} M_1(u):=\sup_{1/5<a\le1/2}F_1(u,a),\quad M_2(u):=\sup_{a\in[u,1/5]}F_2(u,a). \end{equation}

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<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, \begin{equation} 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. \end{equation}

So, the infsup in question is \begin{equation} \inf_{0<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)\vee F_2(u,1/5)]. \end{equation} Since $F_1(u,1/5), F_1(u,1/2), F_2(u,1/5)$ are affine in $u$, it is easy to see that $F_1(u,1/5)\vee F_1(u,1/2)< F_2(u,1/5)$ if $0<u\le1/5$. So,
the infsup in question is \begin{equation} \inf_{0<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]. \end{equation} For brevity, let \begin{equation} g(u):=F_2(u,u),\quad d(u):=F_2(u,u)-F_2(u,1/5)=g(u)-F_2(u,1/5). \end{equation} 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 \begin{equation} \inf_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=2.8343\ldots, \end{equation} attained at $u=0.095260\ldots$.

Because (i) $d(u)=g(u)-F_2(u,1/5)$, (ii) $F_2(u,1/5)$ is affine in $u$, and (iii) $g'(u)$ can change the sign only from $-$ to $+$, we see that $d'(u)$ can change the sign only from $-$ to $+$. Also, $d'(13/100)=-6.4029\ldots<0$. So, $d$ is decreasing on $[0,13/100]$, with $d(13/100)=0.10364\ldots>0$. So, $d>0$ on $[0,13/100]$, that is, $F_2(u,u)>F_2(u,1/5)$ if $0<u\le13/100$. Thus, \begin{equation} \inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]= \min_{0<u\le13/100}F_2(u,u)= 2.8343\ldots, \end{equation} attained at $u=0.095260\ldots$. On the other hand, because $F_2(u,1/5)]$ is increasing in $u$, we have \begin{multline} \inf_{13/100<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]\ge\inf_{13/100<u\le1/5}F_2(u,1/5) \\ = F_2(13/100,1/5) =2.9434\ldots>2.8343\ldots =\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]. \end{multline}

Thus, the infsup in question is \begin{multline} \inf_{0<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)] \\ =\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]\bigwedge \inf_{13/100<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]\\ =\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]= 2.8343\ldots, \end{multline} attained at $a=u=0.095260\ldots$.