The Cauchy-Schwarz Inequality for Inner Product Spaces

Theorem 1 (The Cauchy-Schwarz Inequality for Inner Product Spaces): Let $H$ be an inner product space. Then $|\langle x, y \rangle| \leq \langle x, x \rangle^{1/2} \langle y, y \rangle^{1/2}$ for all $x, y \in H$ where equality holds if and only if $x$ is a scalar multiple of $y$.

• Proof: Let $x, y \in H$ and consider the real-valued polynomial:

(1)

\begin{align} p(t) = \langle y, y \rangle t^2 + 2 |\langle x, y \rangle| t + \langle x, x \rangle \end{align}

• Choose $\xi \in \mathbb{C}$ with $|\xi| = 1$ so that:

(2)

\begin{align} \quad \langle x, \xi y \rangle = |\langle x, y \rangle| \end{align}

(3)

\begin{align} \quad \langle x, \xi y \rangle = \overline{\xi} \langle x, y \rangle \end{align}

• Now consider the inner product of $x + t \xi y$ with itself. We have that:

(4)

\begin{align} \quad \langle x + t \xi y, x + t \xi y \rangle &= \langle x, x \rangle + t \overline{\xi} \langle x, y \rangle + t \xi \langle y, x \rangle + t^2 \xi \overline{\xi} \langle y, y \rangle \\ &= \langle x, x \rangle + t \overline{\xi} \langle x, y \rangle + t \xi \overline{\langle x, y \rangle} + t^2 |\xi| \langle y, y \rangle \\ &= \langle x, x \rangle + 2 \mathrm{Re} (t \xi \langle y, x\rangle) + t^2 \langle y, y \rangle\\ &= \langle x, x \rangle + 2t \mathrm{Re} (\overline{\langle x, \xi y \rangle}) + t^2 \langle y, y \rangle \\ &= \langle x, x \rangle + 2t |\langle x, y \rangle| + t^2 \langle y, y \rangle \\ &= p(t) \end{align}

• By the properties of inner products, the inner product of an element with itself is always greater than or equal to $0$. Therefore $p(t) \geq 0$ for all $t \in \mathbb{R}$. Moreover, $p(0) = 0$. This shows that the descriminant of $p$ is such that:

(5)

\begin{align} \quad \left ( 2t |\langle x, y \rangle | \right )^2 - 4t^2\langle y, y \rangle \langle x, x \rangle \leq 0 \end{align}

• So for $t \neq 0$ we have that:

(6)

\begin{align} |\langle x, y \rangle| \leq \langle x, x \rangle^{1/2} \langle y, y \rangle^{1/2} \end{align}

• And the equality above holds if and only if $x + t\xi y = 0$ which shows that $x$ is a scalar multiple of $y$.