Difference between revisions of "Inner product examples"

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(Created page with "==Of continuous functions== Here the space is {{M|\mathcal{C}_\mathbb{C}[a,b]}} - the continuous functions over the interval {{M|[a,b]}} that are complex va...")
 
m (I had the wrong norm! What I wrote wasn't a norm (failed linearlity!))
 
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: (this is simpler then it sounds as for {{M|f\in\mathcal{C}_\mathbb{C}[a,b]}} we really have {{M|1=f(x)=f_r(x)+jf_i(x)}} where {{M|1=j:=\sqrt{-1} }})
 
: (this is simpler then it sounds as for {{M|f\in\mathcal{C}_\mathbb{C}[a,b]}} we really have {{M|1=f(x)=f_r(x)+jf_i(x)}} where {{M|1=j:=\sqrt{-1} }})
 
{{Begin Inline Theorem}}
 
{{Begin Inline Theorem}}
−
* For {{M|f,g\in\mathcal{C}_\mathbb{C}[a,b]}} we define {{MM|1=\langle f,g\rangle:=\sqrt{\int_a^b{f(x)\overline{g(x)}dx} } }}
+
* For {{M|f,g\in\mathcal{C}_\mathbb{C}[a,b]}} we define {{MM|1=\langle f,g\rangle:=\int_a^b{f(x)\overline{g(x)}dx} }}
 
{{Begin Inline Proof}}
 
{{Begin Inline Proof}}
 
'''Proof that this is an inner product:'''
 
'''Proof that this is an inner product:'''
 
# We require that {{M|1=\langle f,g\rangle=\overline{\langle g,f\rangle} }}
 
# We require that {{M|1=\langle f,g\rangle=\overline{\langle g,f\rangle} }}
 
#* Let us start with {{M|\overline{\langle g,f\rangle} }} and show it is equal to {{M|\langle f,g\rangle}}
 
#* Let us start with {{M|\overline{\langle g,f\rangle} }} and show it is equal to {{M|\langle f,g\rangle}}
−
#** {{MM|1=\overline{\langle g,f\rangle}=\overline{\sqrt{\int^b_a{g(x)\overline{f(x)}dx} } } }}
+
#** {{MM|1=\overline{\langle g,f\rangle}=\overline{\int^b_a{g(x)\overline{f(x)}dx} } }}
−
#**: {{MM|1= =\overline{\sqrt{\int_a^b{(g_r(x)+jg_i(x))(f_r(x)-jf_i(x))dx} } } }}
+
#**: {{MM|1==\overline{\int_a^b(g_r(x)+jg_i(x))(f_r(x)-jf_i(x))dx} }}
−
#**: {{MM|1= =\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} } } }}
+
#**: {{MM|1==\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} } }}
−
#**:: Let:
+
#**:* '''Note:''' the terms are arranged alphabetically but otherwise it's a standard expansion
−
#**::* {{MM|1=a=\int_a^b{[g_r(x)f_r(x)+g_i(x)f_i(x)]dx} }}
+
#**: {{MM|1==\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} }}
−
#**::* {{MM|1=b=\int_a^b{[f_r(x)g_i(x)-f_i(x)g_r(x)]dx} }}
+
#**: {{MM|1==\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} }}
−
#**: {{MM|1= =\overline{\sqrt{a+bj} } }}
+
#**: {{MM|1==\int_a^b{(f_r(x)+jf_i(x))(g_r(x)-jg_i(x))dx} }}
−
#**:: Let:
+
#**: {{MM|1==\int_a^b{f(x)\overline{g(x)}dx} }}
−
#**::* {{MM|1=r=\sqrt{a^2+b^2} }}
+
#** {{MM|1==\langle f,g\rangle}}
−
#**::* {{MM|1=\theta=\text{arctan}(\frac{b}{a})}}
+
#* As required, we have shown {{MM|1=\overline{\langle g,f\rangle}=\langle f,g\rangle}}
−
#**:: So now:
+
# Now we require that {{MM|1=\langle \alpha f+\beta g,h\rangle =\alpha\langle f,h\rangle+\beta\langle g,h\rangle}}
−
#**::* {{MM|1=a+bj=re^{j\theta} }}
+
#* As before, we will start with {{MM|1=\langle \alpha f+\beta g,h\rangle}} and show it is equal to {{MM|\alpha\langle f,h\rangle+\beta\langle g,h\rangle}}
−
#**::* and also: {{MM|1=\overline{\langle g,f\rangle}=\overline{\sqrt{re^{j\theta} } }=\overline{\sqrt{\sqrt{a^2+b^2}e^{j\text{ arctan}(\frac{b}{a})} } } }}
+
#** {{MM|1=\langle \alpha f+\beta g,h\rangle=\int_a^b{(\alpha f(x)+\beta g(x))\overline{h(x)}dx} }}
−
#**::*: {{MM|1= =\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)} } } }}
+
#**: {{MM|1==\alpha\int_a^b{f(x)\overline{h(x)}dx}+\beta\int_a^b{g(x)\overline{h(x)}dx} }}
−
#**::*:* (This is why I have defined variables)
+
#** {{MM|1==\alpha\langle f,h\rangle+\beta\langle g,h\rangle}}
−
#**: {{MM|1= =\overline{\sqrt{r}\ e^{\frac{1}{2}j\theta} } }}
+
#* As required, we have shown {{MM|1=\langle \alpha f+\beta g,h\rangle =\alpha\langle f,h\rangle+\beta\langle g,h\rangle}}
−
#**: {{MM|1= =\sqrt{r}\ e^{-\frac{1}{2}j\theta} }}
+
# Now we need to show that {{M|\forall f\in\mathcal{C}_\mathbb{C}[a,b]}} that {{M|\langle f,f\rangle\ge 0}} (and additionally {{MM|1=\langle f,f\rangle =0\iff f=0}} (that is {{M|f}} is the {{M|0}}-vector, the function that maps all to {{M|0}})
−
#**: {{MM|1= =\sqrt{r\ e^{-j\theta} } }}
+
#* Let {{MM|f\in\mathcal{C}_\mathbb{C}[a,b]}} be given
−
#**: {{MM|1= =\sqrt{a-bj} }}
+
#** {{MM|1=\langle f,f\rangle=\int_a^b{f(x)\overline{f(x)}dx} }}
−
#**: {{MM|1= =\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} } }}
+
#**: {{MM|1==\int_a^b{(f_r(x)+jf_i(x))(f_r(x)-jf_i(x))dx} }}
−
#**: {{MM|1= =\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)
+
#**: {{MM|1==\int_a^b{[f_r(x)^2+f_i(x)^2]dx} }}
−
#**: {{MM|1= =\sqrt{\int_a^b{(f_r(x)+jf_i(x))(g_r(x)-jg_i(x)) dx} } }}
+
#**:* '''Note that:'''
−
#**: {{MM|1= =\sqrt{\int_a^b{f(x)\overline{g(x)}dx} } }}
+
#**:*# {{MM|f_r(x)^2\ge 0}} always
−
#** {{MM|1= =\langle f,g\rangle}} - as required
+
#**:*# {{MM|f_i(x)^2\ge 0}} always
−
#* It is shown that {{MM|1=\langle f,g\rangle=\overline{\langle g,f\rangle} }}
+
#**:* Thus the integral is {{M|\ge 0}}
 +
#** So {{MM|\int_a^b{[f_r(x)^2+f_i(x)^2]dx}\ge 0}}
 +
#* As required, we have shown that {{MM|\langle f,f\rangle\ge 0}} always (as {{M|f}} was arbitrary)
 +
* Now I must show that {{MM|1=\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 {{MM|1=f=(:x\mapsto 0)\implies \langle f,f\rangle=0}} which is the easiest part.
 +
** Just be bothered to write it
 +
{{Todo|Finish this off}}
 
{{End Proof}}{{End Theorem}}
 
{{End Proof}}{{End Theorem}}
 +
 +
[[Category:Functional Analysis]]

Latest revision as of 16:16, 11 July 2015

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]

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

TODO: Finish this off

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