Inner product examples

Of continuous functions

Here the space is [ilmath]\mathcal{C}_\mathbb{C}[a,b][/ilmath] - the continuous functions over the interval [ilmath][a,b][/ilmath] that are complex valued.

(this is simpler then it sounds as for [ilmath]f\in\mathcal{C}_\mathbb{C}[a,b][/ilmath] we really have [ilmath]f(x)=f_r(x)+jf_i(x)[/ilmath] where [ilmath]j:=\sqrt{-1}[/ilmath])

• For [ilmath]f,g\in\mathcal{C}_\mathbb{C}[a,b][/ilmath] we define [math]\langle f,g\rangle:=\sqrt{\int_a^b{f(x)\overline{g(x)}dx} }[/math]

---

Proof that this is an inner product:

1. We require that [ilmath]\langle f,g\rangle=\overline{\langle g,f\rangle}[/ilmath]
   • Let us start with [ilmath]\overline{\langle g,f\rangle} [/ilmath] and show it is equal to [ilmath]\langle f,g\rangle[/ilmath]
     • [math]\overline{\langle g,f\rangle}=\overline{\sqrt{\int^b_a{g(x)\overline{f(x)}dx} } }[/math]

       [math]=\overline{\sqrt{\int_a^b{(g_r(x)+jg_i(x))(f_r(x)-jf_i(x))dx} } }[/math]

       [math]=\overline{\sqrt{\int_a^b{[g_r(x)f_r(x)+g_i(x)f_i(x)]dx}+j\int_a^b{[f_r(x)g_i(x)-f_i(x)g_r(x)]dx} } }[/math]

       Let:

       • [math]a=\int_a^b{[g_r(x)f_r(x)+g_i(x)f_i(x)]dx}[/math]
       • [math]b=\int_a^b{[f_r(x)g_i(x)-f_i(x)g_r(x)]dx}[/math]

       [math]=\overline{\sqrt{a+bj} }[/math]

       Let:

       • [math]r=\sqrt{a^2+b^2}[/math]
       • [math]\theta=\text{arctan}(\frac{b}{a})[/math]

       So now:

       • [math]a+bj=re^{j\theta}[/math]
       • and also: [math]\overline{\langle g,f\rangle}=\overline{\sqrt{re^{j\theta} } }=\overline{\sqrt{\sqrt{a^2+b^2}e^{j\text{ arctan}(\frac{b}{a})} } }[/math]

         [math]=\overline{\sqrt{\sqrt{\left[\int_a^b{[g_r(x)f_r(x)+g_i(x)f_i(x)]dx}\right]^2+\left[\int_a^b{[f_r(x)g_i(x)-f_i(x)g_r(x)]dx}\right]^2}\ e^{j\text{ arctan}\left(\frac{\int_a^b{[f_r(x)g_i(x)-f_i(x)g_r(x)]dx} }{\int_a^b{[g_r(x)f_r(x)+g_i(x)f_i(x)]dx} }\right)} } }[/math]

         • (This is why I have defined variables)

       [math]=\overline{\sqrt{r}\ e^{\frac{1}{2}j\theta} }[/math]

       [math]=\sqrt{r}\ e^{-\frac{1}{2}j\theta}[/math]

       [math]=\sqrt{r\ e^{-j\theta} }[/math]

       [math]=\sqrt{a-bj}[/math]

       [math]=\sqrt{\int_a^b{[g_r(x)f_r(x)+g_i(x)f_i(x)]dx}-j\int_a^b{[f_r(x)g_i(x)-f_i(x)g_r(x)]dx} }[/math]

       [math]=\sqrt{\int_a^b{[f_r(x)g_r(x)+f_i(x)g_i(x)]dx}+j\int_a^b{[f_i(x)g_r(x)-f_r(x)g_i(x)]dx} } \text{ }[/math] (Just by moving the terms around)

       [math]=\sqrt{\int_a^b{(f_r(x)+jf_i(x))(g_r(x)-jg_i(x)) dx} }[/math]

       [math]=\sqrt{\int_a^b{f(x)\overline{g(x)}dx} }[/math]

     • [math]=\langle f,g\rangle[/math] - as required
   • It is shown that [math]\langle f,g\rangle=\overline{\langle g,f\rangle}[/math]