2009-6.md

course	course_year	question_number	tags	title	year
Analysis II	IB	6	IB 2009 Analysis II	Paper 4, Section II, E	2009

Let $(X, d)$ be a metric space with at least two points. If $f: X \rightarrow \mathbb{R}$ is a function, write

$$\operatorname{Lip}(f)=\sup _{x \neq y} \frac{|f(x)-f(y)|}{d(x, y)}+\sup _{z}|f(z)|$$

provided that this supremum is finite. $\operatorname{Let} \operatorname{Lip}(X)={f: \operatorname{Lip}(f)$ is defined $}$. Show that $\operatorname{Lip}(X)$ is a vector space over $\mathbb{R}$, and that Lip is a norm on it.

Now let $X=\mathbb{R}$. Suppose that $\left(f_{i}\right){i=1}^{\infty}$ is a sequence of functions with $\operatorname{Lip}\left(f{i}\right) \leqslant 1$ and with the property that the sequence $f_{i}(q)$ converges as $i \rightarrow \infty$ for every rational number $q$. Show that the $f_{i}$ converge pointwise to a function $f$ satisfying $\operatorname{Lip}(f) \leqslant 1$.

Suppose now that $\left(f_{i}\right){i=1}^{\infty}$ are any functions with $\operatorname{Lip}\left(f{i}\right) \leqslant 1$. Show that there is a subsequence $f_{i_{1}}, f_{i_{2}}, \ldots$ which converges pointwise to a function $f$ with $\operatorname{Lip}(f) \leqslant 1$.