LEC ANA2 3

Completeness

Definition 163.1(Definition).

1. A Cauchy space is a complete metric space.
2. A Banach space is a complete normed linear space .
3. A Hilbert space is a complete inner product space.

Highlights

• $B(S)$ with the sup norm is always a Banach space for any set $S$.
• $C_b(S)$ is a subspace of $B(S)$. It is also a Banach space. When $X$ is compact, $C_b(S)=C(S)$, and $C(S)$ is a Banach space.

Completeness is not a topological property; it depends on the metric. Two metrics may generate induce the same topology (generate the same open sets), but the space may be complete under one metric and not complete under the other. For example, both of the metrics

$$d_1(n,m)=|n-m|,$$ $$d_2(n,m)=\left|1/n-1/m\right|.$$

induce the discrete topology on $\mathbb{N}$. A sequence converges under $d_1$ iff it is constant for all $n$ greater than some $N$. The same holds for $d_2$ (check this!), so a sequence converges under $d_1$ iff it converges under $d_2$.

Under $d_1$, all Cauchy sequences also satisfy the ‘constant after some $N$‘. However, under $d_2$, the sequence $\{x_n=n\}$ is Cauchy, but it doesn’t converge! So $(\mathbb{N},d_1)$ is complete, while $(\mathbb{N},d_2)$ is not.

Exercise 163.2(Exercise).

Let $(X,d)$ be a metric space. Show that $\rho =d/(1+d)$ is a metric. Show that $\{x_n\}$ is Cauchy wrt $d$ $⟺$ $\{x_n\}$ is Cauchy wrt $\rho$.

Example 163.3(Completeness of$B(S)$).

Let $S$ be any set. Define $B(S)$ as we do here, with the metric induced by the sup norm. We will show that $B(S)$ is complete.

Let $\{f_n\}\subseteq B(S)$ be Cauchy. Then,

$$\underset{s\in S}{\sup }|f_n(s)-f_m(s)|\to 0\text{ as }n,m\to 0.$$

Thus, for every $s\in S$, $\{f_n(s){\}}_{n=1}^{\infty }$ is Cauchy in $\mathbb{R}$. Define $f(s):={\lim }_{n\to \infty }{f}_n(s)$. First, we have to show $f\in B(S)$ (that is, $f$ is bounded). Cauchy sequences are bounded; since $B(S)$ is an NLS, we can write $\Vert f_n\Vert \le M$ for some $M$. Then,

$$\begin{array}{rl}|f(s)|& =\underset{n\to \infty }{\lim }|f_n(s)|\\ & \le \underset{n\to \infty }{\lim }\left(|f_n(s)-f_{n^{\prime }}(s)|+|f_{n^{\prime }}(s)|\right)\\ & \stackrel{!}{\le }\underset{n\to \infty }{\lim }\Vert f_n-f_{n^{\prime }}\Vert +\Vert f_{n^{\prime }}\Vert \\ & \le ϵ+\Vert f_{n^{\prime }}\Vert \le ϵ+M\forall s\end{array}$$

so $f\in B(S)$. The limit in $(!)$ exists because $\Vert f_n-f_{n^{\prime }}\Vert$ is a Cauchy sequence¹ in $\mathbb{R}$. We now need to show $f_n\to f$ under $\Vert \cdot {\Vert }_{\infty }$, that is, $\Vert f_n-f\Vert \to 0$. For $ϵ>0$, take $N$ such that $\Vert f_m-f_n\Vert <ϵ$ for all $n,m\ge N$. Then, for all $n\ge N$,

$$\begin{array}{rl}|f_n(s)-f(s)|& =\underset{m\to \infty }{\lim }|f_m(s)-f_n(s)|\forall s\\ & \le \underset{m\to \infty }{\lim }\Vert f_m-f_n\Vert \le ϵ\\ ⟹\Vert f_n-f\Vert & \le ϵ.\end{array}$$

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A good idea is to prove uniform convergence first; boundedness then is immediate. See Pugh (2015) 4.3.

Note that this also shows $l_{\infty }=B(\mathbb{N})$ is complete.

Also note that $B(S)$ with the sup norm is always a Banach space for any set $S$ (no topology or metric needed at all). The structure of $S$ starts to matter only when we restrict to subspaces like $C_b(S)$, because you need a topology on $S$ to know what “continuous” means, and compactness determines whether “continuous implies bounded” (in which case we just write $C(S)$).

Proposition 163.4(Proposition).

If $(X,d)$ is a complete metric space, then $A\subseteq X$ is complete iff $A$ is closed in $X$.

Example 163.5(Example).

Let $C[0,1]\subseteq B[0,1]$ is the space of all continuous functions on $[0,1]$ equipped with the metric induced by the sup norm. We have shown that $C[0,1]$ is complete: essentially, we know that any Cauchy sequence in $C[0,1]$ converges in the complete ambient space $B[0,1]$; showing that the limit lies in $C[0,1]$ requires us to show that the limit is continuous, which is true because the uniform limit of continuous functions is continuous.

Now, consider $C^1[0,1]\subseteq C[0,1]$, the set of all continuously differentiable functions on $[0,1]$. $C^1[0,1]$ is a dense subset of $C[0,1]$², and thus is not complete. However, $C^1[0,1]$ with $\Vert f{\Vert }_{\infty }^1:=\Vert f{\Vert }_{\infty }+\Vert f^{\prime }{\Vert }_{\infty }$ is complete. If $\{f_n\}\subseteq C^1[0,1]$ is Cauchy wrt $\Vert \cdot {\Vert }_{\infty }^1$, then $\{f_n\}$ and $\{f_n^{\prime }\}$ are both Cauchy in $C[0,1]$, and thus must converge to functions $f$ and $g$. We can now say that $f^{\prime }=g$ by Theorem 155.10.

Remark 163.6(Remark).

For a general metric space $S$, you can still consider $C_b(S)$, the space of all bounded continuous functions with the sup norm. This is complete as well; the same proof works. Thus, $C_b(S)$ is a Banach space for all metric spaces $S$. If $S$ happens to be compact, then $C(S)=C_b(S)$ is compact.

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Example 163.7(Example).

Consider the spaces $(C[0,1],\Vert \cdot {\Vert }_1)$ and $(C[0,1],\Vert \cdot {\Vert }_{\infty })$. For every $f\in C[0,1]$,

$$\Vert f{\Vert }_1={\int }_0^1|f|\le \Vert f{\Vert }_{\infty }\cdot 1,$$

so $\Vert \cdot {\Vert }_1\le \Vert \cdot {\Vert }_{\infty }$ on $C[0,1]$. In particular, convergence in $\Vert \cdot {\Vert }_{\infty }$ implies convergence in $\Vert \cdot {\Vert }_1$. Equivalently, by the sequence criterion for continuity, the identity map

$$I:(C[0,1],\Vert \cdot {\Vert }_{\infty })\to (C[0,1],\Vert \cdot {\Vert }_1)$$

is continuous. So, if $S\subseteq C[0,1]$ is closed under $\Vert \cdot {\Vert }_1$, it is closed under $\Vert \cdot {\Vert }_{\infty }$.

The norms are not equivalent, however: there is no constant $c>0$ such that $c\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_1$ for all $f\in C[0,1]$, the classic counterexample being the sequence of triangular ‘spike’ functions of height $1$ and base $1/n$: $\Vert f{\Vert }_{\infty }=1$ for every $n$, but $\Vert f_n{\Vert }_1\to 0$. $(C[0,1],\Vert \cdot {\Vert }_1)$ is not complete, either: $(x^n)\to 1_{\{1\}}\notin C[0,1]$.

Footnotes

1. $\Vert f_n-f_{n^{\prime }}\Vert -\Vert f_m-f_{n^{\prime }}\Vert \le \Vert f_n-f_m\Vert$. ↩

2. Because polynomials are dense in $C[0,1]$, and every polynomial is in $C^1[0,1]$. Proof uses the Weierstrass approximation theorem. ↩

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References

Pugh, C. C. (2015). Real Mathematical Analysis. Springer International Publishing. https://doi.org/10.1007/978-3-319-17771-7