# Extrema of a vector norm under two inner-product constraints.

If $\langle\vec{A},\vec{V}\rangle=1\; ,\; \langle\vec{B},\vec{V}\rangle=c$, then: \begin{align} max\left \| \vec{V} \right \|_{1}=?\;\;\;min\left \| \vec{V} \right \|_{1}=? \end{align} Consider the entry-wise norm.
All of the arrays are positive.
Express the extrema in terms of $\vec{A}$, $\vec{B}$ and $c$.

• Do you mean $\langle A, V\rangle$? – Owen Sizemore Aug 3 '13 at 11:36
• Yes, I exactly mean that (entry-wise product). I will correct it right now. – Amir Kazemi Aug 3 '13 at 11:40
• Without additional information about $A$ and $B$ you can't conclude anything about $\|V\|_1$ other that $\|V\|_1\neq 0$. – Owen Sizemore Aug 3 '13 at 11:47
• Please look at the question below, I tried to ask this question in discrete form. With MATLAB you can find the answers.: math.stackexchange.com/q/457500/88652 – Amir Kazemi Aug 3 '13 at 11:49
• And please notice that you are allowed to response in terms of A and B. – Amir Kazemi Aug 3 '13 at 11:52

You can't get any control on the max. Consider $A=(1, 0)$ and $B=(c, 0)$ then any $V=(1, r)$ for $r\in\mathbb{R}^+$ satisfies this, and $\|V\|_1$ of such a $V$ can be arbitrarily large.
min$\|V\|_1=max\left\{\frac{1}{\|A\|_2}, \frac{c}{\|B\|_2}\right\}$. To see this first recall that $\|V\|_2=\sqrt{\langle V, V\rangle}$, then note that for all vectors, $V$, of a given $\|\cdot\|_2$-norm, the one with the largest value of $\langle A, V\rangle$ are the ones of the form $kA$ where $k$ is a scalar. So to minimize the norm that satisfies $\langle A, V\rangle=1$ we choose $V=\frac{1}{\|A\|_2}A$, a similar argument works with $B$. So you get that $\|V\|_2$ has to be bigger than both of those terms in the max. Since $\|V\|_2\leq\|V\|_1$, you get the same bound on $V$. Choosing A, B, and V as in the above example shows that this bound is realized and thus is the min.