[Short Notes] Non-compactness of the closed unit ball in an infinite-dimensional Banach space

This is about an exercise in [Bass]:

Exercise 19.5. Prove that if H is infinte-dimensional, that is, it has no finite basis, then the closed unit ball in H is not compact.

Proof. Choose an orthonormal basis \{x_i\}, then ||x_i-x_j||^2=||x_i||^2+||x_j||^2=2. This means the sequence is not Cauchy hence has no convergent subsequence.

For a Banach space, by Riesz’s lemma to find a non-Cauchy sequence.


[Bass] Bass, R. F. (2013). Real analysis for graduate students. Createspace Ind Pub.

Lusin’s Theorem and Continuous Extension

Here we give proofs for two versions of Lusin’s Theorem, one from Exercise 44, Ch2 in Folland’s Real Analysis and the other from the textbook used for my first year undergraduate mathematical analysis course in Beijing.  The latter version is a stronger result which in addition discusses the condition for a real-valued function defined on a subset of \mathbb{R}^n to be extended to the whole of \mathbb{R}^n. A more general result in topology is the Tietze Extension Theorem. 

See the full post here: Lusin’s Theorem and Continuous Extension

Here we let \mu denote the Lebesgue measure on \mathbb{R}.

Lusin’s Theorem (Version 1)[Exercise 2.44, Folland]. Suppose E\subset \mathbb{R}^n is Lebesgue  measurable, f: E\to \mathbb{R} is Lebesgue measurable and \epsilon> 0, there is a compact set F\subset E such that \mu(F^c)<\epsilon and f|_F is continuous.

Lusin’s Theorem(Version 2)[Huan]. Suppose E\subset \mathbb{R}^n is Lebesgue measurable and f: E\to \bar{\mathbb{R}} is a Lebesgue measurable extended real valued function with \mu(|f|=+ \infty)=0, then  \forall \epsilon >0, \exists g\in C(E) such that \mu(f\neq g)<\epsilon, where C(E) denotes the space of continuous function on E

Continuous Extension Theorem[Huan]. Suppose E\subset \mathbb{R}^n, then f can be extended to a continuous function on \mathbb{R}^n if and only if f can be extended to a continuous function on the closure \bar{E} of E.

Tietze Extension Theorem. Let X be normal and F \subset X be closed and let f: F \to R be continuous. Then there is a map g: X \to R such that
g(x) = f(x) for all x\in F. (Note that in topology, by a map we mean a continuous function. )