Cauchy-Schwarz inequality

Statement

Let ${ (X,\langle \cdot,\cdot \rangle) }$ be a inner product space. Then,

\(\lvert \langle x,y \rangle \rvert \le \lVert x \rVert \lVert y \rVert\) and the equality holds iff ${ \alpha x + \beta y = \mathbf{0} }$ for some scalars ${ \alpha,\beta }$ s.t. ${ \alpha \beta \neq 0 }$.

The inequality called Cauchy-Schwarz inequality or Cauchy-Bunyakovsky-Schwarz (CBS) inequality

proof

If ${ y=\mathbf{0} }$,

\(\lvert \langle x,0 \rangle \rvert = \lVert x \rVert \lVert \mathbf{0} \rVert\) Suppose ${ y \neq \mathbf{0} }$, and let

\[z = x - \frac{\langle x,y \rangle}{\langle y,y \rangle}y\]

Then,

\[\langle z,y \rangle = \langle x,y \rangle - \langle x,y \rangle = 0\]

Therefore ${ z \perp y }$ and

\[x= \frac{\langle x,y \rangle}{\langle y,y \rangle }y + z\]

which gives

\[\lVert x \rVert^{2} = \left\lvert \frac{\langle x,y \rangle}{\langle y,y \rangle}\right\rvert^{2}\lVert y\rVert^{2} + \lVert z \rVert^{2} = \frac{\lvert \langle x,y \rangle\rvert^{2}}{\lVert y \rVert^{2}} + \lVert z \rVert^{2} \ge \frac{\lvert \langle x,y \rangle\rvert^{2}}{\lVert y \rVert^{2}}\]

Therefore,

\[\lVert x \rVert^{2} \lVert y \rVert^{2} \ge \lvert\langle x,y \rangle\rvert^{2}\]

The equality holds iff ${ z = \mathbf{0} }$ i.e.,

\[\langle y,y \rangle x-\langle x,y \rangle y =0\]