1
$\begingroup$

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.

enter image description 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.

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
1
1
$\begingroup$

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?

$\endgroup$

You must log in to answer this question.