Metrics on Function Spaces

Section 2.5 Metrics on Function Spaces

Another way of viewing Theorem 2.2.1 is that the \(C(E)\text{,}\) the space of continuous functions on \(E\)---assuming \(E\) to be compact, equipped with the metric

\begin{equation*} \rho_{\text{sup}}(f, g) :=\sup_{E}|f(x)-g(x)| \end{equation*}

is a complete metric space.

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On the space \(\mathcal R (\alpha)\) of Riemann-Stieltjes integrable functions, \(\rho_{\text{sup}}(f, g) :=\sup_{E}|f(x)-g(x)|\) is also well defined and becomes a metric on \(\mathcal R (\alpha)\text{.}\) Theorem 2.2.7 implies that \(\mathcal R (\alpha)\) is a complete metric with this metric.

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However, in applications we often need to work with another "metric" on \(\mathcal R (\alpha)\text{:}\)

\begin{equation*} \rho_{L^{1}(\alpha)}(f, g) :=\int_{E}|f(x)-g(x)|\, d\alpha. \end{equation*}

We put a quotation mark on "metric" because it satisfies all the conditions of a metric except for one: \(\rho_{L^{1}(\alpha)}(f, g)=0\) may not imply that \(f=g\) for all \(x\in E\text{.}\) One may use \(\rho_{L^{1}(\alpha)}(f, g)=0\) to define a relation between two functions \(f, g\in \mathcal R (\alpha)\text{,}\) and it’s easy to see that this is an equivalence relation. Let’s continue to use \(\mathcal R (\alpha)\) to denote the space of equivalence classes of \(\mathcal R (\alpha)\) under this equivalence relation.

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Question.

Is \(\mathcal R (\alpha)\) a complete metric space under this metric?

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The answer turns out to be negative in general, and this turns out a major drawback of Riemann integrable. The main advantage of Lebesgue’s integral is that it corrects this deficiency. We will prove that \(C([a,b])\) is dense in \(\mathcal R (\alpha)\) in \(\rho_{L^{1}(\alpha)}\text{.}\) In Lebesgue’s integration theory, it is established that the completion of \(C([a,b])\) in \(\rho_{L^{1}(\alpha)}\) is the space of Lebesgue integrable functions.

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Example 2.5.1. A metric on \(\bbR^{\bbN}\).

Note that \(d(x, y) :=\min\{1, |x-y|\}\) defines a metric on \(\bbR\text{,}\) which makes \(\bbR\) a complete metric space with bounded diameter---the latter follows because \(d(x, y)\le 1\) for all \(x, y\in \bbR\text{.}\) This metric defines the same topology on \(\bbR\) as the usual Euclidean metric does, namely, a set \(U\subset \bbR\) is open in this metric iff it is open in the usual Euclidean metric.

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We can define a metric \(\rho(f, g) :=\sum_{m=1}^{\infty}\min\{1, |f(m)-g(m)|\}/2^{m}\) for \(f, g: \bbN\mapsto \bbR\text{.}\) A sequence \(f_{n}:\bbN\mapsto \bbR\) converges to \(f:\bbN\mapsto \bbR\) pointwise iff \(\rho(f_{n}, f)\to 0\) as \(n\to \infty\text{.}\)

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If we take \(f_{n}(m)=n\) if \(m=n\text{;}\) and \(=0\) if \(m\neq n\text{.}\) Then \(f_{n}(m)\to 0\) pointwise, but not uniformly over \(\bbN\text{,}\) and \(\sum_{m=1}^{\infty}f_{n}(m)=n \to \infty\) as \(n\to \infty\text{.}\)

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Exercise 2.5.2.

Prove the assertion that a sequence \(f_{n}:\bbN\mapsto \bbR\) converges to \(f:\bbN\mapsto \bbR\) pointwise iff \(\rho(f_{n}, f)\to 0\) as \(n\to \infty\text{.}\)

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Exercise 2.5.3.

Prove that the metric space \((\bbR^{\bbN}, \rho)\text{,}\) where \(\rho\) is the metric introduced in the above example, is not compact, but for any \(M > 0\text{,}\) the set \(\{f\in \bbR^{\bbN}: |f(m)|\le M\; \forall m \} \) is a compact set in this metric space.

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