The Schwarz Inequality ! Let ��� be square-integrable functions. Then

��� !!! Proof:

Let f,g be as above. Define the following integrals:

��� !Clearly the first integral vanishes only if f vanishes for all x, in which case the Swartz inequality is saturated. We can therefore consider only non-zero values of this integral. !Define the following function, h: ! ��� !For any function we know that ! ��� !Plugging in the definition of h we get: !

��� From which the Schwarz Inequality follows. !

Q.E.D.

f, g : R ! C

Z 1

�1dx |f(x)|2

Z 1

�1dx0 |g(x0)|2 �

����Z 1

�1dx f(x)⇤g(x)

����2

Iff =

Z 1

�1dxf⇤(x)f(x) Ifg =

Z 1

�1dxf⇤(x)g(x)

h(x) = g(x)� Ifg

Ifff(x)

Z 1

�1dx |h(x)|2 � 0

Z 1

�1dx |h(x)|2 =

Z 1

�1dx (g⇤(x)�

I

⇤fg

Ifff

⇤(x))(g(x)� Ifg

Ifff(x))

=

Z 1

�1dx

|g(x)|2 � Igff

⇤(x)g(x) + Ifgg⇤(x)f(x)

Iff+

IgfIfg

I

2ff

f

⇤(x)f(x)

!

= Igg �|Ifg|2

Iff

� 0