Inner product examples

Of continuous functions

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

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

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

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Proof that this is an inner product:

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

       $$=\overline{\sqrt{\int_a^b{(g_r(x)+jg_i(x))(f_r(x)-jf_i(x))dx} } }$$

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

       Let:

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

       $$=\overline{\sqrt{a+bj} }$$

       Let:

       • $$r=\sqrt{a^2+b^2}$$
       • $$\theta=\text{arctan}(\frac{b}{a})$$

       So now:

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

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

         • (This is why I have defined variables)

       $$=\overline{\sqrt{r}\ e^{\frac{1}{2}j\theta} }$$

       $$=\sqrt{r}\ e^{-\frac{1}{2}j\theta}$$

       $$=\sqrt{r\ e^{-j\theta} }$$

       $$=\sqrt{a-bj}$$

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

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

       $$=\sqrt{\int_a^b{(f_r(x)+jf_i(x))(g_r(x)-jg_i(x)) dx} }$$

       $$=\sqrt{\int_a^b{f(x)\overline{g(x)}dx} }$$

     • $$=\langle f,g\rangle$$ - as required
   • It is shown that $$\langle f,g\rangle=\overline{\langle g,f\rangle}$$