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Suppose that $u\in\mathbb{R}^{n\times 1}$ is such that $\sum_{i=1}^n u_i=0$.

Prove that $|u^\top v|\leq||u||_ 1(\frac{v_{max}-v_{min}}2)$, where $v_{max}=\max |v_i|$ and $v_{min}=\min|v_i|$ for any $v \in \mathbb{R}^{n\times 1}$.

I tried to prove this similarly to how my professor did it, using the fact that $|u^\top(v-\alpha e)|=|u^\top v|$ if $e=(1,\cdots,1)$ and $\alpha$ is any real number because the sum of the components of $u$ is $0$. Then I used Hölder's inequality to get $|u^\top v|\leq||u||_1||v-\alpha e||_\infty$, and this is where I got stuck.

My professor does not justify why $$\inf ||v-\alpha e||_\infty =\frac{v_{max}-v_{min}}2.$$ I guess that the average of the maximum and minimum values of $v$ could minimize $||\cdot||_\infty$, but I'm not really convinced without a proper explanation and I can't prove it myself.

I guess what we really need to find is $$\lim_{p\to\infty} \inf_\alpha (\sum_{i=1}^n |v_i - \alpha|^p)^\frac1p $$ (because the limit of the p-norms is $||\cdot||_\infty$), which does not sound so easy and straightforward, though maybe it actually is and I'm not seeing how.

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Note that $$ v_{\min}-\alpha \leq v_i-\alpha \leq v_\max-\alpha, $$and thus $$ \max_i |v_i -\alpha|= \max\{|v_\min -\alpha|, |v_\max - \alpha|\}. $$ The right-hand side is the maximum of $d(v_\min, \alpha)$ and $d(v_\max, \alpha)$. One can see that the minimizer of it is $$\alpha =\frac{v_\min + v_\max}{2}$$ and $$ \inf_{\alpha\in \mathbb{R}} \|v - \alpha e\|_\infty = \frac{v_\max - v_\min}{2} $$ follows.

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  • $\begingroup$ Oh, I get it now. We want to pick an $\alpha$ that makes both terms we want to find the maximum of be the same will minimize the maximum. Thanks for the explanation! $\endgroup$ – AstlyDichrar Jan 2 at 22:19

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