📂Hilbert Space

Inner Product Spaces and the Cauchy-Schwarz Inequality

Theorem¹

Let $(H, \langle \cdot ,\cdot \rangle)$ be an inner product space. Then, the following inequality holds, and it is called the Cauchy-Schwarz inequality.

$$\left| \langle x,y \rangle \right| \le \langle x,x \rangle^{1/2} \langle y,y \rangle ^{1/2},\quad \forall x,y \in H$$

Explanation

Since a norm can be defined from the inner product, it can also be expressed as the following equation.

$$\left| \left\langle x, y \right\rangle \right| \le \left\| x \right\| \left\| y \right\|,\quad \forall x,y\in H$$

Proof

• Case 1. $x=0$ or $y=0$

  Without loss of generality, let it be $x=0$. Then, by the definition of the inner product,

  $$\left| \langle 0,y \rangle \right| = \left| \langle 0x,y \rangle \right| =0\left| \langle x,y\rangle \right|=0$$

  Thus, it holds.

• Case 2. If $x\ne0$, $y\ne0$ and $\langle x,y \rangle \in \mathbb{R}$

  By the definition of the inner product,

  $$\begin{align*} 0 \le& \left\langle x-\frac{\langle x,y \rangle}{\langle y,y \rangle} y, x-\frac{\langle x,y \rangle}{\langle y,y \rangle} y \right\rangle \\ =&\ \langle x,x \rangle - \frac{\langle x,y \rangle}{\langle y,y \rangle} \langle x,y \rangle -\frac{\langle x,y \rangle}{\langle y,y \rangle} \langle y,x \rangle +\frac{\langle x,y \rangle^{2}}{\langle y,y \rangle^{2}}\langle y,y \rangle \end{align*}$$

  Since $\langle x,y \rangle \in \mathbb{R}$, then $\langle x,y\rangle=\overline{\langle y,x \rangle}=\langle y,x \rangle$. Therefore,

  $$\begin{align*} 0 \le& \langle x,x \rangle - 2\frac{\langle x,y \rangle}{\langle y,y \rangle} \langle x,y \rangle +\frac{\langle x,y \rangle^{2}}{\langle y,y \rangle^{2}}\langle y,y \rangle \\ =&\ \langle x,x \rangle - 2\frac{\langle x,y \rangle^{2}}{\langle y,y \rangle} +\frac{\langle x,y \rangle^{2}}{\langle y,y \rangle} \\ =&\ \langle x,x \rangle - \frac{\langle x,y \rangle^{2}}{\langle y,y \rangle} \end{align*}$$

  Since $\langle y,y \rangle >0$, multiply both sides to get

  $$\begin{align*} && 0 \le& \langle x,x \rangle \langle y,y \rangle - \langle x,y \rangle ^{2} \\ \implies && \langle x,y \rangle ^{2} \le& \langle x,x \rangle \langle y,y \rangle \\ \implies && \left| \langle x,y \rangle \right| \le& \langle x,x \rangle ^{1/2} \langle y,y \rangle ^{1/2} \end{align*}$$

• Case 3. If $x\ne0$, $y\ne 0$ and $\langle x,y\rangle \in \mathbb{C}$

  Let $\left| \lambda \right| =1$ and choose one $\lambda \in \mathbb{C}$ that satisfies $\lambda \langle x,y \rangle\in [0,\infty)$. Then,

  $$\left| \langle x,y \rangle \right| =\left| \lambda \right| \left| \langle x,y \rangle \right|=\left| \lambda \langle x,y \rangle \right|= \lambda\langle x,y \rangle =\langle \lambda x,y \rangle$$

  Therefore, by Case 2,

  $$\begin{align*} \left| \langle x,y \rangle \right| =&\ \langle \lambda x,y \rangle \\ \le& \langle \lambda x, \lambda x \rangle ^{1/2} \langle y,y \rangle ^{1/2} \\ =&\ \langle x,x\rangle^{1/2}\langle y,y\rangle ^{1/2} \end{align*}$$

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