Inner Product Spaces

An inner product space is a vector space, say $\cal H$ , together with a complex bilinear function $\langle~~,~~\rangle$ having the following properties:

(i)   
$\langle f,g\rangle = \overline{\langle g,f\rangle}$
  where 
$f,g \in\cal H$


(ii)   		 
$\langle f,\alpha_1g_1+\alpha_2g_2\rangle = \alpha_1\langle f,g_1\rangle +\alpha_2\langle f,g_2\rangle$


 		 where $\alpha_1$
 and $\alpha_2$
 are complex numbers

(iii)   		 
$\langle f,f\rangle >0$
 if 
$f\not= \vec 0$


 		 and 
$\langle f,f\rangle =0\Leftrightarrow f=\vec 0$
     .

Comments:

(a) The condition $\langle f,g\rangle = \overline{\langle g,f\rangle}$ is quite necessary, otherwise there would be conflict with (iii). Indeed, if $i=\sqrt{-1}$ , then

$$\langle if,if\rangle = i\langle if,f\rangle = i\overline{\langle f,if \rangle} = i(-i)\overline{\langle f,f\rangle}$$

$$= \langle f,f\rangle >0 \quad .$$

In other words, condition (i) guarantees that the positive definiteness condition (iii) is preserved.

(b) With the help of (i), condition (ii) is equivalent to

$$\langle\alpha_1f_1+\alpha_2f_2,g\rangle = \overline{\alpha}_1\langle f_1, g\rangle +\overline{\alpha}_2\langle f_2,g\rangle\,.$$ (11)

Thus we see that a complex scalar (say, $\alpha_1$ or $\alpha_2$ ) in the first factor of the inner product gets complex conjugated when it gets separated from the inner product as a multiplicative factor. One says that $\langle~~~,~~~\rangle$ is linear in the second argument and antilinear in the first argument.

(c) The square root of $\langle f,f\rangle$ , $\sqrt{\langle f,f\rangle}\equiv \Vert f\Vert$ , is called the norm of the vector $f$ . It is always understood that the norm is finite. In particular $\langle f,f\rangle<\infty$

(d) The inner product satisfies the Cauchy-Schwarz inequality

$$\vert\langle f,g\rangle\vert\le\Vert f\Vert\,\Vert g\Vert ~.$$

This inequality has a nice geometrical interpretation for real inner product spaces. In that case $\langle f, g \rangle =\langle g, f \rangle$ is the familiar inner product and

$$-1\le \frac{\langle f, g \rangle }{\Vert f\Vert \Vert g\Vert} \equiv \cos (\hbox{angle between $f$\ and $g$}) \le 1 ~~.$$

The Cauchy-Schwarz inequality follows from the fact that for any complex $\lambda$

$$0\le\langle\lambda f+g,\lambda f+g\rangle = \vert\lambda\vert^2\Vert f\Vert^2 +\Vert g\Vert^2$$

$$+ \overline{\lambda}\langle f,g\rangle +\lambda\langle g,f\rangle ~.$$

Letting $\lambda =x\frac{\langle f,g\rangle}{\vert\langle f,g\rangle\vert}$ we obtain for all real $x$

$$0\le x^2\Vert f\Vert^2+2x\vert\langle f,g\rangle\vert +\Vert g\Vert^2\quad .$$

Consequently, the discriminant,

$$\vert\langle f,g\rangle\vert^2-\Vert f\Vert^2\Vert g\Vert^2\,,$$

of this quadratic expression must be negative or zero, otherwise this expression would be negative for some values of $x$ . It follows that

$$\vert\langle f,g\rangle\vert\le\Vert f\Vert\,\Vert g\Vert\,.$$

(e) The inner product implies the triangle inequality

$$\Vert f\pm g\Vert \le \Vert f\Vert +\Vert g\Vert ~~.$$ (12)

This inequality readily follows from the properties of the inner product (Why?)