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Suppose $a_1,a_2,...,a_n,b_1,b_2,...,b_n$ are real numbers. Does there exist a constant $K$ (depending on $n$ maybe, but not on $a_i$ or $b_i$) such that $(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\leq K(a_1b_1+a_2b_2+...+a_nb_n)^2$?

I don't know how to proceed about it. Any approach/idea is appreciated.

I am basically trying to see whether the well-known Cauchy Schwarz inequality can be reversed in some way. Actually, I am interested in the following problem: does a bound on $|u'v|$ imply any bound on $||u||.||v||$? Here $u,v$ are vectors. This is equivalent to the one I mentioned.

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4 Answers 4

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It is impossible: if the vectors $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ are orthogonal (for the standard inner product), that would imply the vectors are $0$.

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It can be reversed in the following way:

Assume $0 < a_\min \le a_i \le a_\max$ and $0< b_\min \le b_i \le b_\max$, then $\|a\|\cdot\|b\|\le K\langle a,b\rangle$, where $K = \tfrac{1}{2}(c + c^{\texttt{-}1})$ and $c = \sqrt{\frac{a_\max \cdot b_\max}{a_\min\cdot b_\min}}$

This is known as Pólya-Szegö’s inequality.

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  • $\begingroup$ Do you know a necessary and sufficient condition for it to be an equality ? $\endgroup$
    – P. Quinton
    Jul 19, 2021 at 11:51
  • $\begingroup$ @P.Quinton if $a$ and $b$ are both the same constant vector then we have equality; if this is also a necessary condition for equality I don't know for sure, but I would suspect so. $\endgroup$
    – Hyperplane
    Jul 19, 2021 at 12:22
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No, because for example if $a_1=1=b_2$, $b_1=0=a_2$, you would have $$ (1+0)(0+1) \leqslant K(0+0)^2, $$ which is false.

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The answer is NO. For example, $(a_1,a_2)=(1,0)$ and $(b_1,b_2)=(0,1)$.

However, if the vectors $(a_1,\ldots,a_n),(b_1,\ldots,b_n)$, instead of being arbitrary, LIVE is a cone of angle $0\le\vartheta<\pi/4$, then $$ (a_1,\ldots,a_n)\cdot(b_1,\ldots,b_n)\ge \cos(2\vartheta\,)(a_1^2+\cdots a_n^2)^{1/2}(b_1^2+\cdots b_n^2)^{1/2} $$

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