Cauchy-Schwarz Converse - Ratio is required to be bounded

Question

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 ?

Answer 1

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: $$ \exists c,C >0, \forall k,\qquad c a_k \leq b_k \leq C a_k $$