Last modified 2017-09-25 00:25:11 CDT

Proof of the Cauchy-Schwarz Inequality

\[{\large \text{Cauchy-Schwarz Inequality}}\] \[{\lvert \langle f, g \rangle \rvert}^2 \leq \langle f, f \rangle \langle g, g \rangle\]

For two vectors \( f \) and \( g \) in an inner product space, pick an arbitrary real scalar \( \lambda \) and write:

\[\lvert \langle \lambda f + g, \lambda f + g \rangle \rvert \geq 0\]

which is true because it’s a length squared, so

\[\lvert \langle \lambda f + g, \lambda f + g \rangle \rvert = {\lambda}^2 \langle f, f \rangle + 2\lambda\lvert\langle f, g \rangle\rvert + \langle g, g \rangle \geq 0\]

The expression on the left is quadratic in \( \lambda \), and since it’s always greater than or equal to zero, it has zero real roots (i.e. entirely above the x axis) or one real root (i.e. just touching the x axis).

It can’t have two real roots, because that would require that the expression be negative for some \( \lambda \) (i.e. go underneath the x axis), and that can’t be the case since it’s a length squared.

Altogether, this means that the quadratic’s discriminant is less than or equal to zero:

\[b^2 - 4ac \leq 0\] \[4 {\lambda}^2 {\lvert\langle f, g \rangle\rvert}^2 - 4 {\lambda}^2 \langle f, f \rangle \langle g, g \rangle \leq 0\] \[4 {\lambda}^2 {\lvert\langle f, g \rangle\rvert}^2 \leq 4 {\lambda}^2 \langle f, f \rangle \langle g, g \rangle\] \[{\lvert\langle f, g \rangle\rvert}^2 \leq \langle f, f \rangle \langle g, g \rangle\]

Equality occurs when \( f \) and \( g \) are linearly dependent. This means \( f = \lambda g \) for some scalar \( \lambda \), and we can show this directly by writing:

\[{\lvert \langle f, g \rangle \rvert}^2 ={\lvert\langle \lambda g, g \rangle \rvert}^2 = {\lvert \lambda \rvert}^2 {\langle g, g \rangle}^2 = {\lvert \lambda \rvert}^2 \langle g, g \rangle \langle g, g \rangle = \langle f, f \rangle \langle g, g \rangle\] \[{\lvert \langle f, g \rangle \rvert}^2 = \langle f, f \rangle \langle g, g \rangle\]

See some similar/alternate proofs and applications of the Cauchy-Schwarz Inequality here:


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