Recall the Cauchy-Schwarz inequality,
For arbitrary real numbers and
we have
The claim is then that the equality sign holds if and only if there is a real number such that
for each
.
Proof. (





Then, considering the equation and defining,
We have,
But, since we know (by assumption), we have
which is in
(since
since
for at least one
and each term
in nonnegative, so the sum is strictly positive).
() Assume there exists
such that
for each
. Then,
. So,
The forward implication is actually false. Consider 0 as the a-sequence and 9 as the b-sequence. Then the equality holds but there is no real x which will make ax + b = b = 9 = 0 true.
I think it might be a mistake in the book. Because the proof technique in the second implication would still work if there is a y s.t. ak + ybk =0.. this would also imply the Cauchy-Schwartz equality which means there’s something’s wrong with the only if since having such an x isn’t the only way to arrive at the equality. Notice if we change the statement to include the bit about the y, his forward implication is true… If sequence a is all zeros, there exists a y st ak + ybk= 0 namely y=0. So make the iff statement require either there is such an x or there is such a y.
Yess.. I think the statement should be “The equality sign holds if and only if there are real numbers x and y s.t. xak + ybk = 0 for all k.