Notes:Functional Analysis II

Official notes: 8/2/2017 edition

Conventions

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

Chapter 1

Banach spaces: norms and separability

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

Lemma 1.1: -

TODO: Make own lemma

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

Then we have:

• If is a convex set then $N$ is a norm on $X$

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