Administrivia

  • The GSI will be Nick Brody. Office hours and office
  • Midterm by vote will be Monday, Oct 29
  • Course webpage has been updated. Homework is recommended to be written in Latex, submitted on paper
  • Shoutout to arXiv.org - preprints, a wonderful way to see the current state of a subject

Seminorm

As an example of a seminorm, let \(V\) be a vector space over \(\mathbb{R}\), and take \(\varphi \in V'\), a linear functional in the dual space. Define

\[ \| \nu \|_\varphi = p_\varphi(\nu) \]

More than just \(0\) can be in the kernel of \(\varphi\) (trivially take \(\varphi = 0\)), so this is a seminorm.

Sequences

Let \(X\) be a set. A sequence in \(X\) is a function \(f \colon \mathbb{N} \to X\). We notate it as \(\{ x_n \}\) or \(\{ x_n \}_{n \in \mathbb{N}}\).

Let \((X, d)\) be a metric space, and let \(\{x_n\} \subset X\). We say that \(\{ x_n \}\) converges to \(x_n\), written \(\{ x_n \} \to x\), if

\[\forall \epsilon > 0 \ \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, d(x, x_n) < \epsilon\]

Proposition. Let \((X, d_x), (Y, d_y)\) be metric spaces, and let \(f \colon X \to Y\) be continuous. Let \(\{ x_n \} \to x \in X\). Then \(\{ f(x_n) \} \to f(x)\).

This proposition also has a converse. If we apply \(f\) to all sequences converging at a point \(x \in X\) and the resultant sequences also converge, then \(f\) is continuous at that point.

Heading towards completions, let us define some notions in metric spaces.

Let \((X, d)\) be a metric space. Let \(S \subset X\), we say that \(S\) is dense in \(X\) if

\[ \forall x \in X \ \forall \epsilon > 0 \ \exists s \in S \text{ with } d(x, s) < \epsilon \]

For \(x \in X\), the open ball about \(x\) of radius \(\epsilon > 0\) is

\[B(x, \epsilon) = \{ x' \in X \colon d(x, x') < \epsilon \} \]

Proposition. Let \((X, d_x), (Y, d_y)\) be metric spaces. Let \(S \subset X\), with \(S\) dense in \(X\). Let \(f, g \colon X \to Y\) be continuous. Then if \(f|_S = g|_S\), we have \(f = g\).
Proof. Omitted.

For \((X, d)\) a sequence \(\{ x_n \} \subset X\) is a Cauchy sequence if

\[ \forall \epsilon > 0 \ \exists N \in \mathbb{N} \text{ s.t. for } m, n \geq N, d(x_m, x_n) < \epsilon \]

Cauchy sequences are preserved under uniformly continuous maps:

Proposition. Let \((X, d_x), (Y, d_y)\) be metric spaces, and let \(f \colon X \to Y\). If \(f\) is uniformly continuous, and if \(\{ x_n \}\) is a Cauchy sequence in \(X\), then \(\{ f(x_n) \}\) is Cauchy in \(Y\).
Proof. Let \(\epsilon > 0\) be given. Then there is \(\delta > 0\) such that if \(x, x' \in X\) with \(d_x(x, x') < \delta\) then \(d_y(f(x), f(x')) < \epsilon\). Since \(\{ x_n \}\) is Cauchy, there exists an \(N\) s.t. if \(m, n \geq N\), then \(d_x(x_m, x_n) < \delta\). But then \(d_y(f(x_n), f(x_m)) < \epsilon\).

Toward Completions

In an ideal metric space, we want all Cauchy sequences to converge:

Let \(X\) be a metric space. If every Cauchy sequence in \(X\) converges, then \((X, d)\) is said to be complete.

Let \(\mathbb{Q}\) be the set of rational numbers. We can define a metric \(d\) on \(\mathbb{Q}\) by \(d(r, s) = |r - s|\). We intuitively know that \(\mathbb{Q}\) is not complete, as for example \(p(t) = t^2 - 2\) has no roots in \(\mathbb{Q}\).

However, \(\mathbb{R}\) with the same metric is complete, and \(\mathbb{Q}\) is dense in \(\mathbb{R}\).

Inspired by the relationship by \(\mathbb{Q}\) and \(\mathbb{R}\), let’s define a formal notion of completion (draw diagram):

Let \((X, d)\) be a metric space. By a completion of \((X, d)\) we mean a complete metric space \((\overline{X}, d)\) with an isometric map \(f \colon X \to \overline{X}\) with dense range.

Existence of completions is important, but it turns out we can say something about uniqueness of completions by constructing isomorphisms:

Proposition. If \((Y_1, d_{y_1}), f_1 \colon X \to Y_1\) and \((Y_2, d_{y_2}), f_2 \colon X \to Y_2\) are completions of \((X, d)\), then there is a \(g \colon Y_1 \to Y_2\) isometric onto with \(f_2 = g \circ f_1\).