Proof of the Cauchy-Schwarz Inequality
Cauchy-Schwarz Inequality |⟨f,g⟩|2≤⟨f,f⟩⟨g,g⟩For two vectors f and g in an inner product space, pick an arbitrary real scalar λ and write:
|⟨λf+g,λf+g⟩|≥0which is true because it’s a length squared, so
|⟨λf+g,λf+g⟩|=λ2⟨f,f⟩+2λ|⟨f,g⟩|+⟨g,g⟩≥0The 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:
b2−4ac≤0 4λ2|⟨f,g⟩|2−4λ2⟨f,f⟩⟨g,g⟩≤0 4λ2|⟨f,g⟩|2≤4λ2⟨f,f⟩⟨g,g⟩ |⟨f,g⟩|2≤⟨f,f⟩⟨g,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|2⟨g,g⟩2=|λ|2⟨g,g⟩⟨g,g⟩=⟨f,f⟩⟨g,g⟩ |⟨f,g⟩|2=⟨f,f⟩⟨g,g⟩See some similar/alternate proofs and applications of the Cauchy-Schwarz Inequality here: http://cnx.org/content/m10757/latest/