Cauchy Schwarz Inequality

Theorem

Let $ \left( V, \langle \cdot \rangle \right)$ be an inner product space over $\R$.

Then for all $u, v \in V$,

$\displaystyle |\left\langle u, v \right\rangle| \leq \| u \| \cdot \| v \|$

Proof

The assumption that $V$ is real allows the following attractive proof, which utilises the full symmetry of the inner product.

Let $\lambda \in \R$.

The for any $u, v \in V$, since $\langle u, v \rangle \in \R$, we have $\langle u, v \rangle = \langle v, u \rangle$.

Let

	\(\displaystyle \)	\(\displaystyle f_{u,v} ( \lambda )\)	\(=\)	\(\displaystyle \left \langle u - \lambda v, u - \lambda v \right \rangle\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left \langle u, u \right \rangle - 2 \lambda \left \langle u, v \right \rangle + \lambda^2 \left \langle v, v \right \rangle\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lambda^2 \left \Vert v \right \Vert^2 - 2 \lambda \left \langle u, v \right \rangle + \left \Vert u \right \Vert^2\)	\(\displaystyle \)

Letting $a = \| v \|^2$, $b = 2\langle u, v \rangle$, $c = \| u \|^2$, $f_{u,v}$ is the quadratic polynomial.

$\displaystyle f_{u,v}(\lambda) = a\lambda^2 + b\lambda + c$

Furthermore, $f_{u,v}(\lambda) \geq 0$, so $f_{u,v}$ has at most one distinct real root.

Therefore, the discriminant $\Delta$ satisfies

$\displaystyle \Delta = b^2 - 4ac \leq 0$

Therefore,

$\displaystyle 4 \langle u, v \rangle ^ 2 - 4 \| u \|^2 \| v \|^2 \leq 0$

and, dividing through by $4$,

$\displaystyle \langle u, v \rangle ^ 2 \leq \| u \|^2 \| v \|^2$

Taking square roots we have

$\displaystyle |\langle u, v \rangle| \leq \| u \| \| v \|$

$\blacksquare$