Normed and Banach Spaces — Examples

For each fixed coordinate $n$, the scalar sequence $(x_n^{(m)}{)}_m$ is Cauchy because $\mathrm{\mid }{x}_n^{(m)}-x_n^{(\mathrm{\ell })}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Let $x_n:={\lim }_m{x}_n^{(m)}$.

Since $(x^{(m)})$ is Cauchy, choose $M$ with $\mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}<1$ for $m,\mathrm{\ell }\ge M$. Fixing $\mathrm{\ell }=M$ and letting $m\to \mathrm{\infty }$ coordinatewise gives $\mathrm{\mid }{x}_n\mathrm{\mid }\le \mathrm{\parallel }{x}^{(M)}{\mathrm{\parallel }}_{\mathrm{\infty }}+1$ for every $n$, so $x=(x_n)$ is bounded and lies in ${\mathrm{\ell }}^{\mathrm{\infty }}$.

Given $\epsilon >0$, choose $M$ with $\mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$ for $m,\mathrm{\ell }\ge M$. Letting $\mathrm{\ell }\to \mathrm{\infty }$ coordinatewise yields $\mathrm{\mid }{x}_n^{(m)}-x_n\mathrm{\mid }\le \epsilon$ for every $n$, so $\mathrm{\parallel }{x}^{(m)}-x{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon$ for $m\ge M$.

Thus every supremum-norm Cauchy sequence in ${\mathrm{\ell }}^{\mathrm{\infty }}$ converges in ${\mathrm{\ell }}^{\mathrm{\infty }}$, so ${\mathrm{\ell }}^{\mathrm{\infty }}$ is Banach by [L1].

Fix $\epsilon >0$ and choose $m$ with $\mathrm{\parallel }x-x^{(m)}{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}2$. Because $x^{(m)}\in c_0$, there is $N$ with $\mathrm{\mid }{x}_n^{(m)}\mathrm{\mid }<\epsilon \mathrm{/}2$ for all $n\ge N$.

For $n\ge N$, $\mathrm{\mid }{x}_n\mathrm{\mid }\le \mathrm{\mid }{x}_n-x_n^{(m)}\mathrm{\mid }+\mathrm{\mid }{x}_n^{(m)}\mathrm{\mid }<\epsilon \mathrm{/}2+\epsilon \mathrm{/}2=\epsilon$. Hence $x_n\to 0$, so $x\in c_0$.

Step 2.1 shows that $c_0$ is closed in ${\mathrm{\ell }}^{\mathrm{\infty }}$. Since ${\mathrm{\ell }}^{\mathrm{\infty }}$ is Banach by [L1], [L2] makes $c_0$ Banach.

For each $x\in X$, the scalar sequence $(f_m(x){)}_m$ is Cauchy because $\mathrm{\mid }{f}_m(x)-f_{\mathrm{\ell }}(x)\mathrm{\mid }\le \mathrm{\parallel }{f}_m-f_{\mathrm{\ell }}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Define $f(x):={\lim }_m{f}_m(x)$.

Choose $M$ with $\mathrm{\parallel }{f}_m-f_{\mathrm{\ell }}{\mathrm{\parallel }}_{\mathrm{\infty }}<1$ for $m,\mathrm{\ell }\ge M$. Fixing $\mathrm{\ell }=M$ and letting $m\to \mathrm{\infty }$ pointwise gives $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\parallel }{f}_M{\mathrm{\parallel }}_{\mathrm{\infty }}+1$ for every $x$, so $f$ is bounded.

Given $\epsilon >0$, choose $M$ with $\mathrm{\parallel }{f}_m-f_{\mathrm{\ell }}{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$ for $m,\mathrm{\ell }\ge M$. Letting $\mathrm{\ell }\to \mathrm{\infty }$ pointwise yields $\mathrm{\parallel }{f}_m-f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon$ for $m\ge M$. Thus $f_m\to f$ uniformly, and [L1] makes $f$ continuous.

So every supremum-norm Cauchy sequence in $C_b(X)$ converges in $C_b(X)$, and [L2] shows that $C_b(X)$ is Banach.

If $f\in C(K)$, [L2] makes $f$ bounded. Hence $C(K)=C_b(K)$ as sets and the supremum norms agree.

Since $K$ is nonempty and compact, [L1] makes $C_b(K)$ Banach for the supremum norm. By step 1.1, this is exactly $C(K)$ with the same norm.

If $x\in c_0$, then $x^{(N)}\in c_{00}$ and $\mathrm{\parallel }x-x^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{n>N}\mathrm{\mid }{x}_n\mathrm{\mid }\to 0$. So $c_{00}$ is dense in $c_0$ for the supremum norm.

For $1\le p<\mathrm{\infty }$ and $x\in {\mathrm{\ell }}^p$, the truncations satisfy $\mathrm{\parallel }x-x^{(N)}{\mathrm{\parallel }}_p^p={\sum }_{n>N}\mathrm{\mid }{x}_n{\mathrm{\mid }}^p\to 0$, so $c_{00}$ is dense in ${\mathrm{\ell }}^p$ for the ${\mathrm{\ell }}^p$ norm.

The sequence $u=(1,1\mathrm{/}2,1\mathrm{/}3,\dots )$ lies in $c_0$ but not in $c_{00}$, while its truncations $u^{(N)}\in c_{00}$ converge to $u$ in the supremum norm by step 1.1. Hence $c_{00}$ is not complete for that norm, and since $c_0$ is Banach by [L1], [L2] identifies the supremum-norm completion of $c_{00}$ with $c_0$.

Because ${\mathrm{\ell }}^p$ is Banach by [L3], the uniqueness statement [L2] identifies the ${\mathrm{\ell }}^p$-completion of $c_{00}$ with ${\mathrm{\ell }}^p$.

By [L1], the polynomial algebra $P[a,b]$ is dense in $C([a,b])$, so in particular the continuous function $h(x)=\mathrm{\mid }x-c\mathrm{\mid }$ lies in its supremum-norm closure.

The function $h$ is not a polynomial on $[a,b]$: if a polynomial $p$ agreed with $h$, then on $[c,b]$ it would satisfy $p(x)=x-c$, and on $[a,c]$ it would satisfy $p(x)=c-x$. So the polynomial $p(x)-(x-c)$ would vanish on the interval $[c,b]$, hence be identically zero, and similarly $p(x)-(c-x)$ would be identically zero, forcing $x-c=c-x$ for all $x$, impossible because $a<b$.

Steps 1.1 and 1.2 show that $P[a,b]$ is dense and proper in the Banach space $C([a,b])$, so it is not complete.

Since $C([a,b])$ is Banach by [L2] and contains $P[a,b]$ densely, [L3] identifies the completion of $P[a,b]$ with $C([a,b])$.

The first two bullets are exactly the content of [L1].

The third bullet is exactly the content of [L2].

Combining steps 1.1 and 1.2 gives the stated dictionary.

By [L2], the subspace $c_{00}$ is incomplete.

By [L2], the same subspace is dense in $c_0$, so if it were closed then it would equal all of $c_0$. But $c_{00}\mathrm{\ne }{c}_0$, since for example $(1,1\mathrm{/}2,1\mathrm{/}3,\dots )\in c_0\setminus c_{00}$.

Therefore $c_{00}$ is an incomplete normed subspace that is not closed, refuting the statement. The ambient space is Banach by [L1], and [L3] explains why no contradiction occurs: [L3] goes in the opposite direction.

By [L1], the metric $d^{\mathrm{\prime }}(x,y):=\min \{d(x,y),1\}$ has the same topology as the original norm metric $d$.

Suppose $d^{\mathrm{\prime }}$ were induced by some norm $M$ on $V$. Choose a nonzero vector $v$; then $M(v)>0$, so [L2] gives $d^{\mathrm{\prime }}(tv,0)=M(tv)=\mathrm{\mid }t\mathrm{\mid }M(v)$ for every scalar $t$.

The right-hand side of step 2.1 is unbounded as $\mathrm{\mid }t\mathrm{\mid }\to \mathrm{\infty }$, but by definition $d^{\mathrm{\prime }}(tv,0)\le 1$ for every $t$. This contradiction shows that $d^{\mathrm{\prime }}$ is not induced by any norm at all.

Hence $d^{\mathrm{\prime }}$ is topologically equivalent to a norm metric while failing even to come from a norm, so it certainly does not come from a norm equivalent to the original one. This refutes the statement.