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

Proof of the Cauchy-Schwarz Inequality

Cauchy-Schwarz Inequality |f,g|2f,fg,g

For two vectors f and g in an inner product space, pick an arbitrary real scalar λ and write:

|λf+g,λf+g|0

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

|λf+g,λf+g|=λ2f,f+2λ|f,g|+g,g0

The expression on the left is quadratic in λ, 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 λ (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:

b24ac0 4λ2|f,g|24λ2f,fg,g0 4λ2|f,g|24λ2f,fg,g |f,g|2f,fg,g

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

|f,g|2=|λg,g|2=|λ|2g,g2=|λ|2g,gg,g=f,fg,g |f,g|2=f,fg,g

See some similar/alternate proofs and applications of the Cauchy-Schwarz Inequality here: http://cnx.org/content/m10757/latest/

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