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Suppose that $0<p<1$, how to solve the problem $$ \min_{0\leq m \leq 1} 99m^p + (1-m)^p $$ I know the solution is $m^\star = 0$ and it also can be verified easily by a plot, but I can not prove it mathematically.

Can any one help me to prove this problem or give me some hint?

Any help would be highly appreciated, thanks.

enter image description here This is the plot with $p = 0.5$, where x-axis is the value of $m$ and y-axis is the objective function value of this problem.


I tried to solve this problem by KKT condition as follows:

The Lagrangian of this problem is $$ L(m,\lambda_1,\lambda_2) = 99m^p + (1-m)^p -\lambda_1m + \lambda_2(m-1) $$

KKT condition:

\begin{align} 99pm^{p-1} - p(1-m)^{p-1} - \lambda_1 + \lambda_2 &= 0,\\ -m &\leq 0,\\ m-1 &\leq 0,\\ \lambda_1 &\geq 0,\\ \lambda_2 &\geq 0,\\ \lambda_1m &= 0,\\ \lambda_2(m-1) &= 0. \end{align} With the optimal solution $m^\star = 0$, which implying that $\lambda_2 = 0$, $\lambda_1 \geq 0$ but this contradicts with $(0-p) \geq 0$. Also I'm suspected about whether we could use gradient (or KKT) in this problem.

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You are minimizing a $f(m) = 99m^p + (1-m)^p$ over a closed interval. So the boundary points are automatic candidates, so you must consider $f(0) = 1$ and $f(1) = 99$.

Inside the interval you have $$ \begin{split} f'(x) &= 99pm^{p-1} - p(1-m)^{p-1}\\ f''(x) &= 99p(p-1) m^{p-2} + p(p-1)(1-m)^{p-2} \end{split} $$ and since $p-1 < 0$, you have $f''(x) < 0$ everywhere, which means the function is everywhere concave down and all extema inside are maxima. Consequently, there is no candidates for the minima inside, all comes from the boundary.

Thus, you so $f$ over $[0,1]$ is at a minimum when $m=0$ and at a maximum at $m=1$ and is increasing everywhere in between.

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    $\begingroup$ Is the $f'(x) = 99pm^{p-1} - p(1-m)^{p-1}$ or $f'(x) = 99pm^{p-1} + p(1-m)^{p-1}$? I think that the first one is correct, so we cannot know whether $f'(x)$ is positive or not. Thank you for your replying. $\endgroup$
    – Roy Hsu
    Apr 18, 2019 at 3:51
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    $\begingroup$ @RoyHsu fixed thanks $\endgroup$
    – gt6989b
    Apr 18, 2019 at 6:29
  • $\begingroup$ Thank you very much for your answer. It is simple and clear. $\endgroup$
    – Roy Hsu
    Apr 18, 2019 at 6:56

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