$$u^* = \arg\min_{u} \left(\sum_{i=1}^{8} |X(i)-u|^p\right)^{1/p},$$ I need hint solving above equation. I tried deriving with $u$ but I think it's not the right way. where $p\in \{1,\infty\}$, $X(i)=i$
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You need to define your terms. $X(i)$, arg? – herb steinberg Apr 10 '21 at 17:06
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@herbsteinberg X={1,2,3,4,5,6,7,8}, argmin is the value of u which minimises the above equation – anfjnlnl Apr 10 '21 at 17:31
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The question is not clear. From your comment, $X(i)$ takes values from $1,2,...,8$ but $X(i) \ne X(j)$ if $i \ne j$ or not? – NN2 Apr 10 '21 at 18:12
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@NN2 Yes, $X(1)=1,...,X(8)=8$ – anfjnlnl Apr 11 '21 at 05:03
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@username yes $X(i)$ is mentioned above – anfjnlnl Apr 11 '21 at 05:04
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1So $X(i)$ is just $i$? You should edit the clarification into the question, not in the comments. Also, you cannot substitute $p=\infty$ in the expression – Calvin Khor Apr 11 '21 at 05:16
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So you are looking for the minimum of $f$ where $$ f :x\to \left(\sum_{k=1}^8 |k-x|^p\right)^\frac1p. $$ for all $p\in(1,\infty)$ ? The answer is the middle $u=\frac 95$. It is a convex map. – Apr 11 '21 at 07:06
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@username for $p->\infty$ can you write the answer? – anfjnlnl Apr 11 '21 at 09:14
1 Answers
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The problem is equivalent to minimize
$$S = \sum_k^8 |k-u|^p$$
Lemma: for $a,b$ positive and $p>1$ $$a^p+b^p\ge 2^{1-p} |a+b|^p$$ Proof: Applying the Jensen's inequality to the convex function $f(x) = x^p$ (this function is convex because $f''(x) = p(p-1)x^{p-2}>0$ for $p\in (1,+\infty)$):
$$\left|\frac{a}{2} \right|^p+\left|\frac{b}{2} \right|^p\ge \left|\frac{a}{2}+\frac{b}{2} \right|^p$$ or $a^p+b^p\ge 2^{1-p} |a+b|^p$.
The equality occurs if and only if $a=b$ Q.E.D.
Apply the lemma to 4 couples $(a,b) = ((u-k),(8-k-u))$ with $k\in \{1,2,3,4 \}$ $$|k-u|^p+|8-k-u|^p \ge 2^{1-p} |(u-k)+(8-k-u)|^p= 2^{1-p}(8-2k)^p$$
Hence \begin{align} S &= \sum_{k=1}^8 |k-u|^p \\ &= \sum_{k=1}^4 (|k-u|^p+|8-k-u|^p) \ge \sum_{k=1}^4 2^{1-p}(8-2k)^p = 2\sum_{k=1}^4 k^p\\ \end{align}
The equality occurs if and only if $u = \frac{9}{2}$.
Return back to the initial problem, $$\arg\min_{u} \left(\sum_{k=1}^{8} |k-u|^p\right)^{1/p} = \frac{9}{2}$$
NN2
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