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

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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{\int^b_a{g(x)\overline{f(x)}dx} }[/math]

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

       [math]=\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} }[/math]

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

       [math]=\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}[/math]

       [math]=\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}[/math]

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

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

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

       [math]=\alpha\int_a^b{f(x)\overline{h(x)}dx}+\beta\int_a^b{g(x)\overline{h(x)}dx}[/math]

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

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

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

       • Note that:
         1. [math]f_r(x)^2\ge 0[/math] always
         2. [math]f_i(x)^2\ge 0[/math] always
       • Thus the integral is [ilmath]\ge 0[/ilmath]

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

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

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TODO: Finish this off

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