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 $j:=\sqrt{-1}$)

[Expand]

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

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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 $\langle f,g\rangle$
     • $$\overline{\langle g,f\rangle}=\overline{\int^b_a{g(x)\overline{f(x)}dx} }$$

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

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

       • Note: the terms are arranged alphabetically but otherwise it's a standard expansion

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

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

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

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

     • $$=\langle f,g\rangle$$
   • As required, we have shown $$\overline{\langle g,f\rangle}=\langle f,g\rangle$$
2. Now we require that $$\langle \alpha f+\beta g,h\rangle =\alpha\langle f,h\rangle+\beta\langle g,h\rangle$$
   • As before, we will start with $$\langle \alpha f+\beta g,h\rangle$$ and show it is equal to $$\alpha\langle f,h\rangle+\beta\langle g,h\rangle$$
     • $$\langle \alpha f+\beta g,h\rangle=\int_a^b{(\alpha f(x)+\beta g(x))\overline{h(x)}dx}$$

       $$=\alpha\int_a^b{f(x)\overline{h(x)}dx}+\beta\int_a^b{g(x)\overline{h(x)}dx}$$

     • $$=\alpha\langle f,h\rangle+\beta\langle g,h\rangle$$
   • As required, we have shown $$\langle \alpha f+\beta g,h\rangle =\alpha\langle f,h\rangle+\beta\langle g,h\rangle$$
3. Now we need to show that $\forall f\in\mathcal{C}_\mathbb{C}[a,b]$ that $\langle f,f\rangle\ge 0$ (and additionally $$\langle f,f\rangle =0\iff f=0$$ (that is $f$ is the $0$-vector, the function that maps all to $0$)
   • Let $$f\in\mathcal{C}_\mathbb{C}[a,b]$$ be given
     • $$\langle f,f\rangle=\int_a^b{f(x)\overline{f(x)}dx}$$

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

       $$=\int_a^b{[f_r(x)^2+f_i(x)^2]dx}$$

       • Note that:
         1. $$f_r(x)^2\ge 0$$ always
         2. $$f_i(x)^2\ge 0$$ always
       • Thus the integral is $\ge 0$

     • So $$\int_a^b{[f_r(x)^2+f_i(x)^2]dx}\ge 0$$
   • As required, we have shown that $$\langle f,f\rangle\ge 0$$ always (as $f$ was arbitrary)

• Now I must show that $$\langle f,f\rangle =0\implies f=(:x\mapsto 0)$$
  • Use contrapositive to do this, and a bit of analysis
• Finally I must show that $$f=(:x\mapsto 0)\implies \langle f,f\rangle=0$$ which is the easiest part.
  • Just be bothered to write it

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TODO: Finish this off

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