2
$\begingroup$

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?

Thanks in advance!!

$\endgroup$
  • $\begingroup$ What values may $\theta$ take? $\endgroup$ – Iosif Pinelis Jul 24 '18 at 18:12
3
$\begingroup$

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 $a=u=0.099677\ldots$.

$\endgroup$
  • $\begingroup$ Thanks, a lot!! Sorry, I was not precise: $\theta \geq 1$ (so $u\geq 1/5$) and the logarithm is base 2. $\endgroup$ – Armin Weiß Jul 25 '18 at 11:45
  • $\begingroup$ 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). $\endgroup$ – Iosif Pinelis Jul 25 '18 at 15:25
  • $\begingroup$ oops... $\theta \geq 1$ (so $u\leq 1/5$) $\endgroup$ – Armin Weiß Jul 25 '18 at 15:35
  • 1
    $\begingroup$ I have modified the answer, to take into account that $\theta\ge1$ and the log is base $2$. $\endgroup$ – Iosif Pinelis Jul 25 '18 at 17:14
  • $\begingroup$ 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)... $\endgroup$ – Armin Weiß Jul 26 '18 at 14:44

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.