View source for Notes:Functional Analysis II

==Official notes: 8/2/2017 edition==
===Conventions===
* {{M|\mathbb{K} }} - in line with our conventions means either the [[field]] of [[the reals]], {{M|\mathbb{R} }}, or the field of [[complex numbers]], {{M|\mathbb{C} }}
===Chapter 1===
====Banach spaces: norms and separability====
* [[Norm]]
* [[Metric induced by a norm]]
* [[Closed unit ball]] - we will use {{M|\overline{\mathbb{B} } }} or {{M|\overline{\mathbb{B}_X} }} as always. 
** He uses {{M|B_X(0,1)}} - ball notation for {{M|(X,\Vert\cdot\Vert)}} centred at {{M|0}} of radius {{M|1}} - {{Caveat|easily muddled with open ball}}.
* [[Symmetric set (in a vector space)]] - Let {{M|S\in\mathcal{P}(X)}} be given. Symmetric if {{M|\forall x\in S[-x\in S]}}
'''Lemma 1.1:''' - {{XXX|Make own lemma}}
* Suppose that {{M|N:X\rightarrow \mathbb{R} }} satisfies the following three properties:
*# {{M|\forall x\in X[N(X)\ge 0]}},
*# {{M|\forall x\in X[N(x)\eq 0\iff x\eq 0]}}, and
*# {{M|\forall x\in X\forall\lambda\in\mathbb{K}[N(\lambda x)\eq\vert\lambda\vert N(x)]}} {{XXX|Positive definiteness or something right? There's a name for this!}}
: Then we have:
:* If {{M|B:=\eq\{x\in X\ \vert\ N(x)\le 1\} }} is a [[convex set]] then {{M|N}} is a [[norm]] on {{M|X}}
* '''Proof: '''
** We already have 3 of the 4 properties required for {{M|N}} to be a [[norm]], we only need the 4th:
*** {{M|\forall x,y\in X[N(x+y)\le N(x)+N(y)]}} to be done
** With this in mind we start the proof:
** Let {{M|x,y\in X}} be given
*** Suppose {{M|x\eq 0}}, then {{M|N(x+y)\eq N(y)}} and {{M|N(x)\eq 0}} so we see {{M|N(x+y)\eq N(y)\eq N(y)+N(x)}} and this [[implies]] {{M|N(x+y)\le N(x)+N(y)}} as required
*** Suppose {{M|x\neq 0}}, now we have two cases, {{M|y\eq 0}} and {{M|y\neq 0}}:
***# Suppose {{M|y\eq 0}}, then {{M|N(x+y)\eq N(x)}} and {{M|N(y)\eq 0}} so we see {{M|N(x+y)\eq N(y)\eq N(y)+N(x)}} and this [[implies]] {{M|N(x+y)\le N(x)+N(y)}} as required
***# Suppose {{M|y\neq 0}} also
***#* If we want: {{M|N(x+y)\le N(x)+N(y)}} then - after noticing that {{M|N(x)>0}}, {{M|N(x+y)>0}} and {{M|N(y)>0}} - we see this is equivalent to {{MM|\frac{N(x+y)}{N(x)+N(y)}\le 1}} and we notice {{MM|\frac{N(x+y)}{N(x)+N(y)}\eq N\left(\frac{x+y}{N(x)+N(y)}\right)}}
***#** Note also that {{M|y\in B\iff N(y)\le 1}} too! Thus
***#** We really want to show: {{MM|\frac{x+y}{N(x)+N(y)}\in B}}
***#* By basic algebra we see: {{MM|\frac{x+y}{N(x)+N(y)}\eq\frac{1}{N(x)+N(y)}\frac{x}{1}+\frac{1}{N(x)+N(y)}\frac{y}{1} }}
***#* Recall from [[useful field equalities]] that [[1-a/(a+b)=b/(a+b)|{{MM|1-\frac{a}{a+b}\eq \frac{b}{a+b} }}]] so we can attempt to show:
***#** {{MM|\frac{N(x)}{N(x)+N(y)}\frac{x}{N(x)}+\frac{N(y)}{N(x)+N(y)}\frac{y}{N(y)}\in B}} which is just {{MM|\left(1-\frac{N(y)}{N(x)+N(y)}\right)\frac{x}{N(x)}+\frac{N(y)}{N(x)+N(y)}\frac{y}{N(y)}\in B}} {{XXX|Saving work here - rushed this '''carry on from here'''}}
***#* '''Lemma: ''' {{MM|\forall x\in X\left[ n\neq 0\implies \frac{x}{N(x)}\in B\right] }}
***#** Let {{M|x\in X}} be given
***#*** Suppose {{M|x\eq 0}} then by the nature of [[logical implication]] we do not care about the truth or falsity of the RHS and we're done
***#*** Suppose {{M|x\neq 0}}, now we must show {{M|\frac{1}{N(x)} x\in B}}
***#**** However {{M|y\in B\iff N(y)\le 1}} so we want to show the equivalent condition: {{MM|N\left(\frac{x}{N(x)}\right)\le 1}}
***#**** {{MM|N\left(\frac{x}{N(x)}\right)\eq\left\vert\frac{1}{N(x)}\right\vert N(x)\eq \frac{1}{N(x)}N(x)\eq 1 }}
***#***** Thus {{MM|\frac{x}{N(x)}\in B}} - as required
***#** Since {{M|x\in X}} was arbitrary we have shown this holds for all {{M|x\in X}} 
***#* The lemma is shown.
***#* Using the lemma we see: {{MM|\frac{x}{N(x)}\in B}} and {{MM|\frac{y}{N(y)}\in B}}
***#** By [[convexity]] of {{M|B}} we see:
***#*** {{MM|\forall t\in[0,1]\subset\mathbb{R}\left[x+t(y-x)\in B\right] }} and note that {{M|x+t(y-x)\eq (1-t)x+ty}}, thus:
***#*** {{MM|\forall t\in[0,1]\subset\mathbb{R}\left[(1-t)x+ty\in B\right]}}
***#** Hmm