analysis | Parker Glynn-Adey

From Colourings to Fixed Points

The notes for the talk are available here.

Uniform Convergence

The notes for the talk are available here.

Consider $\mathcal{B}$ the Borel $\sigma$ -algebra of Lebesgue measurable subsets of $[0,1]$ . We then define a pseudometric $d(A,B) = \mu(A \Delta B)$ on $\mathcal{B}$ where $A \Delta B = A \setminus B \cup B \setminus A$ is the symmetric difference of sets. Ignoring sets of zero measure, we have that $d$ is indeed a metric. We wish to show that $\mathcal{B}$ is not sequentially compact in its metric.

Write $x = d_0.d_1d_2\dots$ for the binary expansion of $x \in [0,1]$ . Consider $E_i = \{x = d_0.d_1d_2\dots\ :\ d_i = 1\}$ . We compute $E_i \Delta E_j = \{x : d_i \neq d_j\}$ . If we think probabilistically, where the digits $d_1$ and $d_j$ represent independent coin tosses, we get: $\mu(E_i \Delta E_j) = 1/2$ for $i \neq j$ . Thus $d(E_i,E_j)$ is constant for $i \neq j$ and hence $\{E_i\}$ can have no convergent subsequences

This came from the September 2005 UoT Analysis comprehensive. The solution is due to Dror Bar-Natan.

Tagged with: analysis

BPM 1.4.2 (Sp 84)

Posted in Math by pgadey on 2011/03/10

Let $f(x) = x \log(1 + 1/x)$ for $0 < x < \infty$ . We show that $f$ is monotonically increasing. First we rewrite $f(x) = x \log \left( \frac{x + 1}{x} \right) = x [ \log(x+1) - \log(x)]$ . Taking derivatives, and doing some algebra, we get $f'(x) = [\log(x + 1) - \log(x)] - 1/(x+1)$ . We wish to show that $0 < f'(x)$ everywhere. This would follow from $1/(x+1) < \log(x + 1) - \log(x)$ . However, this follows from the convexity of $\log(x)$ . Which says that $1/(x + \epsilon) < [\log(x + \epsilon) - \log(x)] / \epsilon$ for all $0 < x, \epsilon$ .

We compute $\lim_{x \rightarrow 0} f(x)$ . Taking $x_n = 1/(e^n - 1)$ we see that $f(x_n) = 1/(e^n - 1) \log(e^n) = n/(e^n - 1)$ . We then have that $f(x_n) \rightarrow 0$ . Since $f$ is monotonically increasing, and positive, we have that $\lim f(x) = 0$ as $x \rightarrow 0$ . We compute $\lim_{x \rightarrow \infty} f(x)$ . Applying concavity, we have $f(e^n) = e^n \log(1 + e^{-n}) \geq e^n \epsilon e^{-n} = \epsilon$ for all $0 < \epsilon < 1$ . We then check that $f(x) = x \log(1 + 1/x) \stackrel{\star}{\leq} x (1/x) \leq 1$ , where $\star$ is by applying $\log(1 + y) \leq y$ . It follows that $\lim_{x \rightarrow \infty} f(x) = 1$ .