Notes:Functional Analysis II

Official notes: 8/2/2017 edition

Conventions

• [ilmath]\mathbb{K} [/ilmath] - in line with our conventions means either the field of the reals, [ilmath]\mathbb{R} [/ilmath], or the field of complex numbers, [ilmath]\mathbb{C} [/ilmath]

Chapter 1

Banach spaces: norms and separability

• Norm
• Metric induced by a norm
• Closed unit ball - we will use [ilmath]\overline{\mathbb{B} } [/ilmath] or [ilmath]\overline{\mathbb{B}_X} [/ilmath] as always.
  • He uses [ilmath]B_X(0,1)[/ilmath] - ball notation for [ilmath](X,\Vert\cdot\Vert)[/ilmath] centred at [ilmath]0[/ilmath] of radius [ilmath]1[/ilmath] - Caveat:easily muddled with open ball.
• Symmetric set (in a vector space) - Let [ilmath]S\in\mathcal{P}(X)[/ilmath] be given. Symmetric if [ilmath]\forall x\in S[-x\in S][/ilmath]

Lemma 1.1: -

TODO: Make own lemma

• Suppose that [ilmath]N:X\rightarrow \mathbb{R} [/ilmath] satisfies the following three properties:
  1. [ilmath]\forall x\in X[N(X)\ge 0][/ilmath],
  2. [ilmath]\forall x\in X[N(x)\eq 0\iff x\eq 0][/ilmath], and
  3. [ilmath]\forall x\in X\forall\lambda\in\mathbb{K}[N(\lambda x)\eq\vert\lambda\vert N(x)][/ilmath]
     TODO: Positive definiteness or something right? There's a name for this!

Then we have:

• If is a convex set then [ilmath]N[/ilmath] is a norm on [ilmath]X[/ilmath]

• Proof:
  • We already have 3 of the 4 properties required for [ilmath]N[/ilmath] to be a norm, we only need the 4th:
    • [ilmath]\forall x,y\in X[N(x+y)\le N(x)+N(y)][/ilmath] to be done
  • With this in mind we start the proof:
  • Let [ilmath]x,y\in X[/ilmath] be given
    • Suppose [ilmath]x\eq 0[/ilmath], then [ilmath]N(x+y)\eq N(y)[/ilmath] and [ilmath]N(x)\eq 0[/ilmath] so we see [ilmath]N(x+y)\eq N(y)\eq N(y)+N(x)[/ilmath] and this implies [ilmath]N(x+y)\le N(x)+N(y)[/ilmath] as required
    • Suppose [ilmath]x\neq 0[/ilmath], now we have two cases, [ilmath]y\eq 0[/ilmath] and [ilmath]y\neq 0[/ilmath]:
      1. Suppose [ilmath]y\eq 0[/ilmath], then [ilmath]N(x+y)\eq N(x)[/ilmath] and [ilmath]N(y)\eq 0[/ilmath] so we see [ilmath]N(x+y)\eq N(y)\eq N(y)+N(x)[/ilmath] and this implies [ilmath]N(x+y)\le N(x)+N(y)[/ilmath] as required
      2. Suppose [ilmath]y\neq 0[/ilmath] also
         • If we want: [ilmath]N(x+y)\le N(x)+N(y)[/ilmath] then - after noticing that [ilmath]N(x)>0[/ilmath], [ilmath]N(x+y)>0[/ilmath] and [ilmath]N(y)>0[/ilmath] - we see this is equivalent to [math]\frac{N(x+y)}{N(x)+N(y)}\le 1[/math] and we notice [math]\frac{N(x+y)}{N(x)+N(y)}\eq N\left(\frac{x+y}{N(x)+N(y)}\right)[/math]
           • Note also that [ilmath]y\in B\iff N(y)\le 1[/ilmath] too! Thus
           • We really want to show: [math]\frac{x+y}{N(x)+N(y)}\in B[/math]
         • By basic algebra we see: [math]\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} [/math]
         • Recall from useful field equalities that [math]1-\frac{a}{a+b}\eq \frac{b}{a+b} [/math] so we can attempt to show:
           • [math]\frac{N(x)}{N(x)+N(y)}\frac{x}{N(x)}+\frac{N(y)}{N(x)+N(y)}\frac{y}{N(y)}\in B[/math] which is just [math]\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[/math]
             TODO: Saving work here - rushed this carry on from here
         • Lemma: [math]\forall x\in X\left[ n\neq 0\implies \frac{x}{N(x)}\in B\right] [/math]
           • Let [ilmath]x\in X[/ilmath] be given
             • Suppose [ilmath]x\eq 0[/ilmath] then by the nature of logical implication we do not care about the truth or falsity of the RHS and we're done
             • Suppose [ilmath]x\neq 0[/ilmath], now we must show [ilmath]\frac{1}{N(x)} x\in B[/ilmath]
               • However [ilmath]y\in B\iff N(y)\le 1[/ilmath] so we want to show the equivalent condition: [math]N\left(\frac{x}{N(x)}\right)\le 1[/math]
               • [math]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 [/math]
                 • Thus [math]\frac{x}{N(x)}\in B[/math] - as required
           • Since [ilmath]x\in X[/ilmath] was arbitrary we have shown this holds for all [ilmath]x\in X[/ilmath]
         • The lemma is shown.
         • Using the lemma we see: [math]\frac{x}{N(x)}\in B[/math] and [math]\frac{y}{N(y)}\in B[/math]
           • By convexity of [ilmath]B[/ilmath] we see:
             • [math]\forall t\in[0,1]\subset\mathbb{R}\left[x+t(y-x)\in B\right] [/math] and note that [ilmath]x+t(y-x)\eq (1-t)x+ty[/ilmath], thus:
             • [math]\forall t\in[0,1]\subset\mathbb{R}\left[(1-t)x+ty\in B\right][/math]
           • Hmm