Section 2.4 Normed and Inner Product Spaces
Note on the exam.
The statement of the theorem below is very much examinable, as is the general idea of the proof. The full details of the proof are not examinable, except in as much as they are involved in assigned problem sheet questions.
There are analogues of Theorem 2.10 and Theorem 2.11 for normed spaces and for inner product spaces.
Theorem 2.12. Completions of normed and inner product spaces.
- The completion of a normed space is also a normed space, hence a Banach space. The norm is a continuous extension of the original norm.
- The completion of an inner product space is also an inner product space, hence a Hilbert space. The inner product is a continuous extension of the original inner product.
Proof.
Let \((X,\n\blank_X)\) be a normed space. Following the proof of Theorem 2.10, we write \(\seq xn \sim \seq yn\) if \(\lim_{n \to \infty} \n{x_n-y_n}_X = 0\text{,}\) set
\begin{align*}
\hat X \amp = \{ \text{Cauchy sequences in }(X,\n\blank_X) \} / \sim,\\
\hat X_0 \amp = \{ [(x)_{n\in\N}] : x \in X \}\text{,}
\end{align*}
and define a bijection \(f \maps X \to \hat X_0\) by \(f(x) = [(x)_{n \in
\N}]\text{.}\) We then define our norm \(\n\blank_{\hat X}\maps \hat X \to \R\) by
\begin{equation}
\big\|[\seq xn]\big\|_{\hat X} = \lim_{n\to \infty} \n {x_n}_X\text{.}\tag{2.6}
\end{equation}
This is an extension of \(\n\blank_X\) in the sense that \(\n{f(x)}_{\hat X} = \n x_X\) for all \(x \in X\text{.}\)
It is not at all obvious from its definition that \(\hat X\) is a vector space, or even how the relevant operations should be defined. A natural guess, though, it to try and ‘commute’ everything past the equivalence class brackets. That is, for Cauchy sequences \(\seq xn\) and \(\seq yn\) in \((X,\n\blank_X)\) and real numbers \(\alpha \in \R\text{,}\) we define
\begin{equation}
[\seq xn] + [\seq yn] = [(x_n+y_n)_{n\in \N}],
\qquad
\alpha [\seq xn] = [\seq{\alpha x}n].\tag{2.7}
\end{equation}
Since equivalence classes are involved, we must check that these are well defined, in particular that they do not depend on which representatives we choose. We request the details in Exercise 2.4.1 and Exercise 2.4.2.
Now that we know \(\hat X\) is a vector space, we must show that \(\n\blank_{\hat X}\) defined in (2.6) is a norm on \(\hat X\text{.}\) As in the proof of Theorem 2.10, this follows by taking limits of the axioms for \(\n\blank_X\text{.}\) For instance, to see the triangle inequality, let \(\seq xn\) and \(\seq
yn\) be Cauchy sequences in \((X,\n\blank_X)\text{.}\) Then for each \(n \in
\N\text{,}\) the triangle inequality for \(\n\blank_X\) gives
\begin{equation*}
\n{x_n+y_n}_X \le \n{x_n}_X + \n{y_n}_X \text{.}
\end{equation*}
Sending \(n \to \infty\) and using (2.7), we conclude that
\begin{align*}
\big\Vert[\seq xn] + [\seq yn]\big\Vert_{\hat X}
\amp =
\big\Vert[(x_n+y_n)_{n\in\N}]\big\Vert_{\hat X}\\
\amp \le
\big\Vert[\seq xn]\big\Vert_{\hat X}
+ \big\Vert[\seq yn]\big\Vert_{\hat X}\text{.}
\end{align*}
The density of \(\hat X_0\) and the fact that it is isometric to \(X\) follow exactly as in the proof of Theorem 2.10. The proof for inner product spaces is likewise very similar, with the natural definition
\begin{equation*}
\big\langle[\seq xn],[\seq yn]\big\rangle_{\hat X} = \lim_{n\to\infty} \scp{x_n}{y_n}_X\text{.}
\end{equation*}
We omit the details.
Exercises Exercises
1. Vector addition and Cauchy sequences.
Let \(\seq xn\) and \(\seq yn\) be Cauchy sequences in a normed space \((X,\n\blank_X)\text{.}\)
(a)
Show that \((x_n+y_n)_{n \in \N}\) is also a Cauchy sequence.
(b)
Suppose that \(\seq {x'}n\) and \(\seq {y'}n\) are Cauchy sequences in \((X,\n\blank_X)\text{.}\) Show that \(\seq {x'}n \sim \seq xn\) and \(\seq {y'}n \sim \seq yn\) implies \((x_n+y_n)_{n\in\N} \sim
(x'_n+y'_n)_{n\in\N}\text{,}\) where here \(\seq an \sim \seq bn\) means \(\lim_{n\to \infty}\n{a_n-b_n}=0\text{.}\)
2. (PS6) Scalar multiplication and Cauchy sequences.
Let \(\alpha \in \R\) and let \(\seq xn\) be a Cauchy sequence in a normed space \((X,\n\blank_X)\text{.}\)
(a)
Show that \(\seq{\alpha x}n\) is also a Cauchy sequence.
Solution.
If \(\alpha = 0\text{,}\) then \(\seq{\alpha x}n\) is the constant sequence \((0,0,\ldots)\text{,}\) which is clearly convergent and hence Cauchy. So assume that \(\alpha \ne 0\) and let \(\varepsilon \gt 0\text{.}\) Since \(\seq xn\) is Cauchy, we can choose \(N \in \N\) such that, for all \(n,m \ge N\text{,}\) \(\n{x_n-x_m}_X \lt \varepsilon/\abs \alpha\text{.}\) Then, for all \(n,m \ge N\text{,}\) we have
\begin{align*}
\n{\alpha x_n-\alpha x_m}_X
\amp =
\n{\alpha (x_n- x_m)}_X\\
\amp =
\abs \alpha \n{x_n- x_m}_X\\
\amp
\lt \abs \alpha \frac \varepsilon{\abs \alpha}\\
\amp
= \varepsilon\text{.}
\end{align*}
Comment.
In the argument above we want to show that the sequence \((\alpha
x_n)_{n\in\N}\) is Cauchy, and so given an \(\varepsilon \gt 0\) we must choose \(N \in \N\) appropriately based on this \(\varepsilon\text{.}\) This logical relationship between \(\varepsilon\) and \(N\) is at the heart of the definition of a Cauchy sequence, and getting it wrong on an exam (or being overly vague) would probably mean losing most or even all of the marks on a question like this one — even though the resulting argument might superficially look quite similar to the official solution. If you are at all confused about this important point, I encourage you in the strongest possible terms to look back at Exercise B.1 from Problem Sheet 1 as well as all of Appendix G, and also to get in touch with me over email or in to office hours.
(b)
Let \(\seq {x'}n\) be another Cauchy sequence in \((X,\n\blank_X)\text{.}\) Show that \(\seq {x'}n \sim \seq xn\) implies \(\seq{\alpha x'}n \sim
\seq{\alpha x}n\text{,}\) where here \(\seq an \sim \seq bn\) means \(\lim_{n\to \infty}\n{a_n-b_n}_X=0\text{.}\)
Solution.
We simply calculate
\begin{align*}
\lim_{n\to \infty} \n{\alpha x'_n - \alpha x_n}_X
\amp =
\lim_{n\to \infty}\Big( \abs \alpha \n{x'_n - x_n}_X \Big)\\
\amp =
\abs \alpha \lim_{n\to \infty} \n{x'_n - x_n}_X \\
\amp = 0\text{.}
\end{align*}
