Find a $\bar v$ with $\|\bar v\|_\infty=\alpha$ such that $\langle u \, , \bar v \rangle = - \|u\|_1\|v\|_\infty = -\alpha\|u\|_1$

Question

The original question is:

Find a $\bar v \in \mathbb{R}^n$ with $\|\bar v\|_\infty=\alpha$ and $\alpha>0$ express it in terms of $u \in \mathbb{R}^n\setminus\{0\}$ and $\alpha$ such that,

$\langle u \, , \bar v \rangle = - \|u\|_1\|\bar v\|_\infty = -\alpha\|u\|_1$

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The question leading up to this one was to show that,

$\langle u \, , v \rangle \geq - \|u\|_1\|v\|_\infty \quad \forall \,u,v\in\mathbb{R}^n$

which i tred to proove by multiplying with (-1)

$\Leftrightarrow -\langle u \, , v \rangle \leq \|u\|_1\|v\|_\infty$

$\Leftrightarrow \langle -u \, , v \rangle \leq \|u\|_1\|v\|_\infty$

where we can use the Cauchy–Schwarz inequality

$\langle -u \, , v \rangle \leq \|u\|\|v\|$ and by plugging in the 1-Norm $\langle -u \, , v \rangle \leq \|u\|_1\|v\|_1 \leq \|u\|_1\|v\|_\infty$

Answer 1

Let $v_i$ = $\text{sign }u_i$. Now Let $\bar{v} = -\alpha v$. So

$$\langle u, \bar{v} \rangle = \langle u, -\alpha v \rangle = -\alpha\langle u, v \rangle = -\alpha \|u \|_1$$

For the other question (I think your proof is incorrect):

$$\langle -u, v\rangle = \sum_{i=1}^n -u_iv_i \le \sum_{i=1}^n |u_i||v_i| \le \sum_{i=1}^n |u_i| \|v \|_{\infty} = \|u\|_1 \|v\|_{\infty}$$

Answer 2

Since $||v||=\alpha$, there is an $n$ such that $|v_n|=\alpha$ Let $u_n=-\alpha$ and $u_k=0$ for $k\ne n$. Then $(u,v)=-\alpha^2=-\alpha||u||_1$.