Cauchy-Schwarz Converse – Ratio is required to be bounded
cauchy-schwarz-inequalityinequality
I am reading Cauchy-Schwarz master class book, and they deduced that (5.1) cannot be true unless the ratio $\frac{a_k}{b_k}$ is bounded from above and below(page 74).
Can someone explain it to me ?
Best Answer
Allowing the ratio to be unbounded allows for $(a_k)_k,(b_k)_k$ to be orthogonal; to illustrate this, an example would be
$a_1=\dots=a_{n/2}=1$, $b_1=\dots=b_{n/2}=0$ and $a_{n/2+1}=\dots=a_n=0$, $b_{n/2+1}=\dots=b_n=1$. Then, the LHS is strictly positive (in our example, $n/2$) but the RHS is zero, as $\langle a,b\rangle = \sum_{k=1}^n a_kb_k= 0$.
Without being so extreme, you could approach this "near-orthogonality" condition, making any hope of a converse Cauchy—Schwarz hopeless. Intuitively, we have the following idea:
Since Cauchy—Schwarz is true with equality when $a,b$ are colinear, to have a converse to Cauchy—Schwarz we want (and need) "near-colinearity," with is captured by this bounded ratio condition:
$$
\exists c,C >0, \forall k,\qquad c a_k \leq b_k \leq C a_k
$$
$(x+y+z)^2 = \left(x\sqrt{\frac{y+z}{y+z}} +y\sqrt{\frac{x+z}{x+z}}+z\sqrt{\frac{x+y}{x+y}}\right)^2\leq\left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right)(y+z+x+z+x+y)$. You will get it. I didn't read that book, but I believe Srivatsan is right.
Hint: The quadratic equation $ax^2+bx+c=0$ has real roots if and only if $b^2-4ac\ge 0$.
Added: The quadratic polynomial $\sum_1^n (a_ix+b_i)^2$ is a sum of squares. Thus this polynomial is always $\ge 0$.
Recall that a quadratic $ax^2+bx+c$, where $a\gt 0$, is always $\ge 0$ if and only if the discriminant $b^2-4ac$ is $\le 0$. Compute the discriminant of the messy quadratic. The inequality $b^2-4ac\le 0$ turns out to be precisely the C-S Inequality (well, we have to divide by $4$).
As to when we have equality, the quadratic has a real root $k$ if and only if $a_ik+b_i=0$ for all $i$. This is the case iff $b_i=-ka_i$, meaning that the $a_i$ and $b_i$ are proportional.
By the way, things are I think marginally prettier if we look at the polynomial $\sum_1^n (a_ix-b_i)^2$.
Best Answer
Allowing the ratio to be unbounded allows for $(a_k)_k,(b_k)_k$ to be orthogonal; to illustrate this, an example would be $a_1=\dots=a_{n/2}=1$, $b_1=\dots=b_{n/2}=0$ and $a_{n/2+1}=\dots=a_n=0$, $b_{n/2+1}=\dots=b_n=1$. Then, the LHS is strictly positive (in our example, $n/2$) but the RHS is zero, as $\langle a,b\rangle = \sum_{k=1}^n a_kb_k= 0$.
Without being so extreme, you could approach this "near-orthogonality" condition, making any hope of a converse Cauchy—Schwarz hopeless. Intuitively, we have the following idea: