# Using Hölder's inequality to prove statement for a specific vector $u$

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.

## 1 Answer

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.

• 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! – AstlyDichrar Jan 2 at 22:19