Difference between revisions of "Inner product examples"

Latest revision as of 16:16, 11 July 2015

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 $\langle f,g\rangle:=\int_a^b{f(x)\overline{g(x)}dx}$

Proof that this is an inner product:

We require that ⟨f,g⟩=¯⟨g,f⟩
- Let us start with ¯⟨g,f⟩ and show it is equal to ⟨f,g⟩
  - ¯⟨g,f⟩=¯∫bag(x)¯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 $\overline{\langle g,f\rangle}=\langle f,g\rangle$
Now we require that ⟨αf+βg,h⟩=α⟨f,h⟩+β⟨g,h⟩
- As before, we will start with ⟨αf+βg,h⟩ and show it is equal to α⟨f,h⟩+β⟨g,h⟩
  - $\langle \alpha f+\beta g,h\rangle=\int_a^b{(\alpha f(x)+\beta g(x))\overline{h(x)}dx}$
    $=\alpha\int_a^b{f(x)\overline{h(x)}dx}+\beta\int_a^b{g(x)\overline{h(x)}dx}$
  - $=\alpha\langle f,h\rangle+\beta\langle g,h\rangle$
- As required, we have shown $\langle \alpha f+\beta g,h\rangle =\alpha\langle f,h\rangle+\beta\langle g,h\rangle$
Now we need to show that ∀f∈CC[a,b] that ⟨f,f⟩≥0 (and additionally ⟨f,f⟩=0⟺f=0 (that is f is the 0-vector, the function that maps all to 0)
- Let f∈CC[a,b] be given
  - ⟨f,f⟩=∫baf(x)¯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:
    $f_r(x)^2\ge 0$ always
    
    $f_i(x)^2\ge 0$ 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 ⟨f,f⟩=0⟹f=(:x↦0)
- Use contrapositive to do this, and a bit of analysis
Finally I must show that f=(:x↦0)⟹⟨f,f⟩=0 which is the easiest part.
- Just be bothered to write it

TODO: Finish this off

@@ Line 3: / Line 3: @@
 : (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}}
-* 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}}
 '''Proof that this is an inner product:'''
 # 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}}
-#** {{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}}
+[[Category:Functional Analysis]]

Difference between revisions of "Inner product examples"

Latest revision as of 16:16, 11 July 2015

Of continuous functions

Navigation menu

Views

Personal tools

Navigation

Search

Tools