The Cauchy-Schwarz Inequality

# The Cauchy-Schwarz Inequality

We are about to look at one of the more significant inequalities in mathematics known as the Cauchy-Schwarz inequality. Before we do so though, we will first need to be able to write each vector $u$ in an inner product space as a sum of a scalar multiple of a vector $v$ and a vector $w$ that is orthogonal to $v$.

Let $a \in \mathbb{F}$. Then $u = av + u - av = av + (u - av)$. We now want $(u - av)$ to be orthogonal to $v$. This happens if:

(1)\begin{align} \quad 0 = <u - av, v> \\ \quad 0 = <u, v> + <-av, v> \\ \quad 0 = <u, v> -a<v, v> \\ \quad 0 = <u, v> - a \| v \|^2 \\ \quad a \| v \|^2 = <u, v> \\ \quad a = \frac{<u, v>}{\| v \|^2} \end{align}

Thus we have that:

(2)\begin{align} \quad u = av + (u - av) \\ \quad u = \frac{<u, v>}{\| v \|^2} v - \left ( u - \frac{<u, v>}{\| v \|^2} v \right ) \end{align}

We are now ready to state and prove the Cauchy-Schwarz inequality.

Theorem 1 (The Cauchy-Schwarz Inequality): Let $V$ be an inner product space. Then:a) $\mid <u, v> \mid ≤ \| u \| \| v \|$ for all $u, v \in V$.b) $\mid <u, v> \mid = \| u \| \| v \|$ if and only if one of $u$ or $v$ is a scalar multiple of the other. |

**Proof of a)**Let $V$ be an inner product space and let $u, v, w \in V$ and suppose that $v$ is orthogonal to $w$. Suppose that $v = 0$. Then $\mid <u, v> \mid = \mid <u, 0> \mid = 0$ and $\| u \| \| v \| = \| u \| \| 0 \| = 0$ so the inequality holds. Suppose instead that $v \neq 0$. Let $u$ be written as the sum of $v$ and $w$ such that:

\begin{align} \quad u = \frac{<u, v>}{\| v \|^2} v + w \\ \end{align}

- Now we can apply The Pythagorean Theorem for Inner Product Spaces since $u$ is the sum of two vectors $v$ and $w$ which are orthogonal.

\begin{align} \quad \| u \|^2 = \biggr \| \frac{<u, v>}{\| v \|^2} v \biggr \|^2 + \| w \|^2 = \frac{\mid <u, v> \mid}{\| v \|^2} + \| w \| ≥ \frac{\mid <u, v> \mid^2}{\| v \|^2} \end{align}

- We now multiply both sides of this inequality by $\| v \|^2$ and then take the square root of both sides

\begin{align} \quad \| u \|^2 ≥ \frac{\mid <u, v> \mid^2}{\| v \|^2} \\ \quad \| u \|^2 \| v \|^2 ≥ \mid <u, v> \mid^2 \\ \quad \| u \| \| v \| ≥ \mid < u, v> \mid \end{align}

**Proof of b)**$\Rightarrow$ Suppose that $\mid <u, v> \mid = \| u \| \| v \|$. Then from the Cauchy-Schwarz inequality proof above we must have that:

\begin{align} \quad \| u \|^2 = \frac{\mid <u, v> \mid^2}{\| v \|^2} + \| w \|^2 \end{align}

- But then this implies that $\| w \|^2 = 0$ and so $w = 0$. But then $u = \frac{<u, v>}{\| v \|^2} v + w = \frac{<u, v>}{\| v \|^2} v$. Therefore $u$ is a scalar multiple of $v$.

- $\Leftarrow$ Suppose that $u$ is a scalar multiple of $v$. Then $u = kv$ for some $k \in \mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$) and so:

\begin{align} \quad \mid <u, v> \mid = \mid <kv, v> \mid = \mid k \mid \mid <v, v> \mid = \mid k \mid \| v \|^2 = \mid k \mid \| v \| \| v || = \| kv \| \| v \| = \| u \| \| v \| \quad \blacksquare \end{align}