# Cauchy-Bunyakovsky-Schwarz inequality

1. Oct 9, 2008

### phyguy321

absolute value( $$\int{fg}$$) $$\leq$$[($$\int{f}$$$$^{2}$$)($$\int{g}$$$$^{2}$$)]$$^{1/2}$$

all integrals are from a to b
does anyone have any idea of a proof for this?

2. Oct 9, 2008

### morphism

3. Oct 16, 2008

### morphism

Bump? What more do you want?

4. Oct 16, 2008

### HallsofIvy

Staff Emeritus
5. Oct 17, 2008

### phyguy321

I dont want to solve it in terms of a vector space, I would like to see it using integrals

6. Oct 17, 2008

### HallsofIvy

Staff Emeritus
Then use the "vector space" proof when the vector space in question is a function space!

The "vector space" proof looks at the inner product of $\vec{u}- \lambda\vec{v}$ with itself, getting $|\vec{u}-\lambda\vec{v}|= <\vec{u}- \lambda\vec{v},\vec{u}= \lambda\vec{v}>= <\vec{u},\vec{u}>- 2\lambda<\vec{u},\vec{v}>+ \lambda^2<\vec{v},\vec{v}>\ge 0$, for all $\lambda$ because it is a "square".

In particular, if we let let
$$\lambda= \frac{<\vec{u},\vec{v}>}{<\vec{v},\vec{v}>}$$
that becomes
$$<\vec{u},\vec{u}>-2\frac{\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}}+ \frac{\vec{u},\vec{v}>^2}{\vec{v},\vec{v}}$$
$$= <\vec{u},\vec{u}>- \frac{<\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}>}\ge 0$$
so
$$<\vec{u},\vec{u}>\ge \frac{<\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}>}$$
and
$$<\vec{u},\vec{u}><\vec{v},\vec{v}>\ge<\vec{u},\vec{v}>^2$$

That is precisely the proof for "vector spaces" at the site I linked to. Now, the set of all integrable functions IS an inner product space with innerproduct given by $<f, g>= \int fg$.

Replacing $<\vec{u},\vec{v}>$ by $\int fg$ , $<\vec{u}, \vec{v}>$ with $\int f^2$, and $<\vec{v}, \vec{v}>$ with $\int g^2$ gives exactly the inequality you post. That's the whole point of "abstraction" in mathematics- you don't have to prove a lot of different versions of the same thing.