# How to prove Cauchy Schwartz Inequality for norms in Lebesgue Integration

I am self studying Apostol ( Mathematical Analysis) but I couldn't prove this particular theorem given in text despite the hint given . 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.

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.

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?