I am self studying Apostol ( Mathematical Analysis) but I couldn't prove this particular theorem given in text despite the hint given .
So, I am asking here.
Its part (e) , I have no idea how to use RHS from the inequality to prove the CS inequality.
Any help will be really appreciated.
On
Cauchy-Schwarz inequality holds for all symetric semi-definite bilinear forms, so for inner products in particular. The proof is general.
Suppose $(\cdot,\cdot)$ is an inner product on a vector space $E$.
Let $x,y \in E$ and consider the polynomial
$$ P(\lambda) = \|x+\lambda y\|^2 = \|x\|^2 + 2 (x,y) \lambda + \|y\|^2 \lambda^2. $$
Suppose $y \neq 0$ (otherwise the inequality is immediate).
This polynomial is of degree 2 and is non-negative on all $\mathbb R$ so its discriminant is non-positive, that is:
$$ (2 (x,y))^2 - 4\|y\|^2\|x\|^2 \leq 0 $$ i.e. $$ |(x,y)| \leq \|x\|\|y\| $$
which is Cauchy-Schwarz inequality.
Does that help?
We have
$$(f(x)g(y)-g(x)f(y))^2=f(x)^2g(y)^2-2f(x)g(x)f(y)g(y)+g(x)^2f(y)^2.$$
It follows
$$ \int_I(f(x)g(y)-g(x)f(y))^2 dy=f(x)^2||g||^2-2f(x)g(x) (f,g)+g(x)^2||f||^2.$$
Hence
$$ \int_I[ \int_I(f(x)g(y)-g(x)f(y))^2dy]dx= ||f||^2 ||g||^2-2(f,g)^2+||g||^2||f||^2.$$
From
$$ \int_I[ \int_I(f(x)g(y)-g(x)f(y))^2dy]dx \ge0$$
we get
$$||f||^2 ||g||^2-2(f,g)^2+||g||^2||f||^2 \ge 0$$
and Cauchy - Schwarz follows.