Normed and Banach Spaces

Since $x=(x-y)+y$, the triangle inequality in [L1] gives $\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }x-y\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }$.

Since $y=(y-x)+x$ and $\mathrm{\parallel }y-x\mathrm{\parallel }=\mathrm{\parallel }x-y\mathrm{\parallel }$ by absolute homogeneity at the scalar $-1$, [L1] also gives $\mathrm{\parallel }y\mathrm{\parallel }\le \mathrm{\parallel }x-y\mathrm{\parallel }+\mathrm{\parallel }x\mathrm{\parallel }$.

Step 1.1 yields $\mathrm{\parallel }x\mathrm{\parallel }-\mathrm{\parallel }y\mathrm{\parallel }\le \mathrm{\parallel }x-y\mathrm{\parallel }$, and step 1.2 yields $\mathrm{\parallel }y\mathrm{\parallel }-\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }x-y\mathrm{\parallel }$; together these are exactly $\mid \mathrm{\parallel }x\mathrm{\parallel }-\mathrm{\parallel }y\mathrm{\parallel }\mid \le \mathrm{\parallel }x-y\mathrm{\parallel }$.

The displayed inequality says precisely that the map $x\mapsto \mathrm{\parallel }x\mathrm{\parallel }$ is $1$-Lipschitz for the metric $d(u,v)=\mathrm{\parallel }u-v\mathrm{\parallel }$.

By [L1], the metric on $W$ induced by the restricted norm is exactly the ambient norm metric restricted to $W$.

The hypothesis that $W$ is complete for its restricted norm therefore says that $W$ is complete as a metric subspace of $V$.

Apply [L2] under [A1]. For each fixed point in the closure, [A1] supplies the approximating sequence; completeness and uniqueness of its ambient limit then put that point in $W$. Thus $W$ is closed in $V$. No simultaneous selection of sequences for all closure points is required.

By [L3], the ambient norm metric on $V$ is complete.

By [L1], the restricted norm on $W$ induces exactly the ambient subspace metric on $W$.

Since $W$ is closed in the complete metric space $V$, [L2] makes that subspace metric complete.

Completeness of the restricted norm metric is exactly the Banach property for $W$ by [L3].

For addition, $\mathrm{\parallel }(x+h)+(y+k)-(x+y)\mathrm{\parallel }=\mathrm{\parallel }h+k\mathrm{\parallel }\le \mathrm{\parallel }h\mathrm{\parallel }+\mathrm{\parallel }k\mathrm{\parallel }$ by [L2].

For scalar multiplication, $(\lambda +\alpha )(x+h)-\lambda x=\lambda h+\alpha x+\alpha h$, so $\mathrm{\parallel }(\lambda +\alpha )(x+h)-\lambda x\mathrm{\parallel }\le \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }h\mathrm{\parallel }+\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }h\mathrm{\parallel }$ by the triangle inequality and absolute homogeneity from [L1] and [L2].

Given $\epsilon >0$, step 1.1 shows that if $\mathrm{\parallel }h\mathrm{\parallel }<\epsilon \mathrm{/}2$ and $\mathrm{\parallel }k\mathrm{\parallel }<\epsilon \mathrm{/}2$, then $(x+h)+(y+k)$ lies in the $\epsilon$-ball around $x+y$; this is continuity of addition at $(x,y)$.

For scalar multiplication at $(\lambda ,x)$, first require $\mathrm{\mid }\alpha \mathrm{\mid }<1$ and then choose $\delta >0$ so that $\mathrm{\mid }\lambda \mathrm{\mid }\delta +\delta (\mathrm{\parallel }x\mathrm{\parallel }+\delta )<\epsilon$; step 1.2 then gives $\mathrm{\parallel }(\lambda +\alpha )(x+h)-\lambda x\mathrm{\parallel }<\epsilon$ whenever $\mathrm{\mid }\alpha \mathrm{\mid }<\delta$ and $\mathrm{\parallel }h\mathrm{\parallel }<\delta$.

Since the point $(x,y)$ was arbitrary, step 2.1 proves continuity of addition everywhere, and step 2.2 proves continuity of scalar multiplication everywhere.

For each coordinate, $\mathrm{\parallel }{x}_k{\mathrm{\parallel }}^2$ is one of the nonnegative summands in ${\sum }_{j<n}\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$, so $\mathrm{\parallel }{x}_k\mathrm{\parallel }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2$ for every $k$ and therefore $\mathrm{\parallel }x{\mathrm{\parallel }}_{\max }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2$.

Applying the scalar inequality $\mathrm{\parallel }(a_k){\mathrm{\parallel }}_2\le \mathrm{\parallel }(a_k){\mathrm{\parallel }}_1$ from [L2] to the nonnegative real tuple $(\mathrm{\parallel }{x}_0\mathrm{\parallel },\dots ,\mathrm{\parallel }{x}_{n-1}\mathrm{\parallel })$ gives $\mathrm{\parallel }x{\mathrm{\parallel }}_2\le \mathrm{\parallel }x{\mathrm{\parallel }}_1$.

Since every coordinate norm is at most $\mathrm{\parallel }x{\mathrm{\parallel }}_{\max }$, summing $n$ such bounds gives $\mathrm{\parallel }x{\mathrm{\parallel }}_1\le n\mathrm{\parallel }x{\mathrm{\parallel }}_{\max }$.

The three displayed inequalities of steps 1.1, 1.2, and 1.3 are exactly the comparison constants needed for equivalence of the three norms.

If $x^{(m)}$ is Cauchy for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\max }$, then each coordinate sequence $(x_k^{(m)}{)}_m$ is Cauchy in $E_k$, because $\mathrm{\parallel }{x}_k^{(m)}-x_k^{(\mathrm{\ell })}\mathrm{\parallel }\le \mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\max }$ for every $k<n$.

Since each $E_k$ is Banach, [L1] gives a point $x_k\in E_k$ with $x_k^{(m)}\to x_k$. Put $x:=(x_0,\dots ,x_{n-1})$.

Given $\epsilon >0$, choose $M_0$ so that $\mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\max }<\epsilon \mathrm{/}2$ for $m,\mathrm{\ell }\ge M_0$, and for each $k<n$ choose $M_k$ so that $\mathrm{\parallel }{x}_k^{(m)}-x_k\mathrm{\parallel }<\epsilon \mathrm{/}2$ for $m\ge M_k$. For $M:=\max \{M_0,M_1,\dots ,M_{n-1}\}$ and $m\ge M$, taking $\mathrm{\ell }\to \mathrm{\infty }$ in a fixed coordinate gives $\mathrm{\parallel }{x}_k^{(m)}-x_k\mathrm{\parallel }\le \epsilon \mathrm{/}2$ for every $k<n$, hence $\mathrm{\parallel }{x}^{(m)}-x{\mathrm{\parallel }}_{\max }\le \epsilon \mathrm{/}2<\epsilon$.

Thus the product is complete for the maximum norm, hence Banach for that norm by [L1].

The inequalities in [L3] show that a sequence is Cauchy or convergent for one standard product norm exactly when it is so for the others. Therefore completeness for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\max }$ is equivalent to completeness for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$ and for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$, so the product is Banach for all three norms.

By [L1] and [L2], for every $\epsilon >0$ there is $N$ such that ${\sum }_{\mathrm{\ell }\le n<m}\mathrm{\parallel }{x}_n\mathrm{\parallel }<\epsilon$ whenever $m>\mathrm{\ell }\ge N$.

For such $m>\mathrm{\ell }\ge N$, the tail of the vector partial sums satisfies $s_m-s_{\mathrm{\ell }}={\sum }_{\mathrm{\ell }\le n<m}{x}_n$.

Applying the triangle inequality to the finite sum in step 1.2 gives $\mathrm{\parallel }{s}_m-s_{\mathrm{\ell }}\mathrm{\parallel }\le {\sum }_{\mathrm{\ell }\le n<m}\mathrm{\parallel }{x}_n\mathrm{\parallel }<\epsilon$.

Since this holds for every $\epsilon >0$, the partial sums $(s_m)$ are Cauchy in the norm metric, which is exactly the claim.

Assume $V$ is Banach. If $\sum x_n$ is absolutely convergent, [L2] makes its partial sums Cauchy, and [L1] then makes those partial sums converge in $V$. So every absolutely convergent series converges.

Assume [A1]. Let $(y_m)$ be a Cauchy sequence in $V$. Choose inductively a strictly increasing sequence $(n_k)$ with $\mathrm{\parallel }{y}_m-y_n\mathrm{\parallel }<2^{-k}$ whenever $m,n\ge n_k$. Then $\mathrm{\parallel }{y}_{n_{k+1}}-y_{n_k}\mathrm{\parallel }<2^{-k}$ for every $k$.

The series ${\sum }_{k=0}^{\mathrm{\infty }}(y_{n_{k+1}}-y_{n_k})$ is absolutely convergent because $\sum 2^{-k}$ converges and each term has norm at most $2^{-k}$. By [A1] it therefore converges to some $z\in V$.

Its partial sums are $y_{n_m}-y_{n_0}$, so the subsequence $(y_{n_m})$ converges to $y_{n_0}+z$.

Given $\epsilon >0$, choose $K$ so that $\mathrm{\parallel }{y}_m-y_n\mathrm{\parallel }<\epsilon \mathrm{/}2$ for $m,n\ge K$, and choose $m$ with $n_m\ge K$ and $\mathrm{\parallel }{y}_{n_m}-(y_{n_0}+z)\mathrm{\parallel }<\epsilon \mathrm{/}2$. Then for every $n\ge K$, $\mathrm{\parallel }{y}_n-(y_{n_0}+z)\mathrm{\parallel }\le \mathrm{\parallel }{y}_n-y_{n_m}\mathrm{\parallel }+\mathrm{\parallel }{y}_{n_m}-(y_{n_0}+z)\mathrm{\parallel }<\epsilon$. So $(y_n)$ converges, and $V$ is complete.

Step 1.1 proves $(1)\Rightarrow (2)$ and steps 1.2 through 4.1 prove $(2)\Rightarrow (1)$, so the two conditions are equivalent.

Termwise sums and scalar multiples of Cauchy sequences are again Cauchy: the triangle inequality gives $\mathrm{\parallel }(x_n+y_n)-(x_m+y_m)\mathrm{\parallel }\le \mathrm{\parallel }{x}_n-x_m\mathrm{\parallel }+\mathrm{\parallel }{y}_n-y_m\mathrm{\parallel }$, and absolute homogeneity gives $\mathrm{\parallel }\lambda x_n-\lambda x_m\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }{x}_n-x_m\mathrm{\parallel }$.

Likewise $\mathrm{\parallel }\lambda x_n-\lambda x_n^{\mathrm{\prime }}\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }{x}_n-x_n^{\mathrm{\prime }}\mathrm{\parallel }$ for every $n$, so [L1] and [L3] give $\rho ((\lambda x_n),(\lambda x_n^{\mathrm{\prime }}))=0$. Hence scalar multiplication on classes is representative-independent.

Because $(x_n)$ is Cauchy, the reverse triangle inequality in [L2] gives $\mid \mathrm{\parallel }{x}_n\mathrm{\parallel }-\mathrm{\parallel }{x}_m\mathrm{\parallel }\mid \le \mathrm{\parallel }{x}_n-x_m\mathrm{\parallel }$; so the real sequence $(\mathrm{\parallel }{x}_n\mathrm{\parallel })$ is Cauchy and therefore convergent by [L3].

If $[x_n]=[x_n^{\mathrm{\prime }}]$ and $[y_n]=[y_n^{\mathrm{\prime }}]$, then $\mathrm{\parallel }(x_n+y_n)-(x_n^{\mathrm{\prime }}+y_n^{\mathrm{\prime }})\mathrm{\parallel }\le \mathrm{\parallel }{x}_n-x_n^{\mathrm{\prime }}\mathrm{\parallel }+\mathrm{\parallel }{y}_n-y_n^{\mathrm{\prime }}\mathrm{\parallel }$ for every $n$; passing to limits and using [L1] and [L3] gives $\rho ((x_n+y_n),(x_n^{\mathrm{\prime }}+y_n^{\mathrm{\prime }}))=0$. So addition on classes is representative-independent.

If $[x_n]=[x_n^{\mathrm{\prime }}]$, then $\mid \mathrm{\parallel }{x}_n\mathrm{\parallel }-\mathrm{\parallel }{x}_n^{\mathrm{\prime }}\mathrm{\parallel }\mid \le \mathrm{\parallel }{x}_n-x_n^{\mathrm{\prime }}\mathrm{\parallel }$ for every $n$; the right-hand side tends to $0$ by [L1], so the two real norm sequences have the same limit by [L3]. Thus $\mathrm{\parallel }[x_n]\mathrm{\parallel }:={\lim }_n\mathrm{\parallel }{x}_n\mathrm{\parallel }$ is well defined.

By [L2], the Cauchy classes carry termwise addition and scalar multiplication, and each class has a well-defined norm $\mathrm{\parallel }[x_n]\mathrm{\parallel }:={\lim }_n\mathrm{\parallel }{x}_n\mathrm{\parallel }$.

On constant sequences these operations agree with those of $X$, and $\mathrm{\parallel }i(x)\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ by definition, so $i$ is a linear isometry.

For classes $[x_n]$ and $[y_n]$, the distance of the published metric completion is ${\lim }_n\mathrm{\parallel }{x}_n-y_n\mathrm{\parallel }$, while the new difference class is $[x_n-y_n]$; therefore the completion metric is exactly the metric induced by the norm from step 1.1.

By [A1], the underlying metric space is complete by [L1], so step 2.2 makes $\widehat{X}$ complete for its new norm metric. Thus $\widehat{X}$ is Banach and $(\widehat{X},i)$ is a completion of $X$ in the sense of [L3].

Suppose a second Banach-space structure on the same underlying set induces the published completion metric and also makes $i$ a dense linear isometry. Its addition and scalar multiplication are continuous by [L4], and they agree with the operations of step 1.1 on the dense subset $i[X]\times i[X]$ and on $\mathbb{K}\times i[X]$; therefore they agree everywhere.

The norm is then forced as well, because in any compatible normed structure $\mathrm{\parallel }u\mathrm{\parallel }$ equals the metric distance from $u$ to $0$. So the compatible Banach-space structure is unique.

For $x,y\in X$, linearity gives $Tx-Ty=T(x-y)$, so the bound in [L2] yields $\mathrm{\parallel }Tx-Ty\mathrm{\parallel }\le C\mathrm{\parallel }x-y\mathrm{\parallel }$. Thus $T$ is Lipschitz and uniformly continuous.

Define $S:i[X]\to Y$ by $S(i(x)):=T(x)$. This is well defined because $i$ is injective as a linear isometry, and $S$ is uniformly continuous with the same constant $C$.

Since $i[X]$ is dense in the Banach space $\widehat{X}$ and $Y$ is complete, [A1] and [L3] give a unique continuous map $\widehat{T}:\widehat{X}\to Y$ with $\widehat{T}\text{\hspace{0.17em}⁣}{\upharpoonright }_{i[X]}=S$, equivalently $\widehat{T}\circ i=T$.

Let $u,v\in \widehat{X}$. Density and [A1] supply sequences $i(x_n)\to u$ and $i(y_n)\to v$ from the dense copy. By [L4], $i(x_n+y_n)=i(x_n)+i(y_n)\to u+v$ in $\widehat{X}$, so continuity of $\widehat{T}$ gives $\widehat{T}(u+v)={\lim }_n\widehat{T}(i(x_n+y_n))={\lim }_{n}T(x_n+y_n)={\lim }_n(T{x}_n+T{y}_n)\mathrm{.}$ Using linearity of $T$ and continuity of addition in $Y$ from [L4], the right-hand side is ${\lim }_{n}T{x}_n+{\lim }_{n}T{y}_n=\widehat{T}(u)+\widehat{T}(v)\mathrm{.}$ Hence $\widehat{T}(u+v)=\widehat{T}(u)+\widehat{T}(v)$.

For $u\in \widehat{X}$, density and [A1] supply a sequence $i(x_n)\to u$ from the dense copy. Continuity gives $\widehat{T}(u)={\lim }_{n}T(x_n)$, and step 1.1 yields $\mathrm{\parallel }T(x_n)\mathrm{\parallel }\le C\mathrm{\parallel }{x}_n\mathrm{\parallel }=C\mathrm{\parallel }i(x_n)\mathrm{\parallel }$ for every $n$; passing to the limit gives $\mathrm{\parallel }\widehat{T}(u)\mathrm{\parallel }\le C\mathrm{\parallel }u\mathrm{\parallel }$. So $\widehat{T}$ is bounded with the same constant.

Let $\lambda$ be a scalar and let $u\in \widehat{X}$, and use [A1] with density to choose a sequence $i(x_n)\to u$ from the dense copy. By [L4], $i(\lambda x_n)=\lambda i(x_n)\to \lambda u$ in $\widehat{X}$, so continuity of $\widehat{T}$ gives $\widehat{T}(\lambda u)={\lim }_n\widehat{T}(i(\lambda x_n))={\lim }_{n}T(\lambda x_n)={\lim }_n\lambda T(x_n)=\lambda {\lim }_{n}T(x_n)=\lambda \widehat{T}(u)\mathrm{.}$ Thus $\widehat{T}$ is homogeneous, and with step 4.1 it is linear.

The continuity extension in step 3.1 was unique, so no second bounded linear map can also extend $T$. Therefore $\widehat{T}$ is the unique bounded linear extension of $T$ across the completion.

By [A1] and [L1], there is a unique isometry $U:Y\to Z$ with $U\circ j=k$.

On the dense subspace $j[X]\subseteq Y$, the map $j(x)\mapsto k(x)$ is linear and norm-preserving. Applying [L2] under [A1] to this dense linear isometry extends it to a bounded linear map $\tilde{U}:Y\to Z$ with $\tilde{U}\circ j=k$.

Both $U$ and $\tilde{U}$ are continuous maps $Y\to Z$ extending the same map on $j[X]$, so the uniqueness in step 1.1 forces $U=\tilde{U}$. Hence $U$ is linear.

Since $U$ is already an isometry by step 1.1, it is a linear isometric isomorphism. Uniqueness of such a map is again the uniqueness from [L1].