Banach-Space Differential Calculus and Banach Manifolds: Examples

Expanding with [L1], $B(x+h,y+k)-B(x,y)=B(h,y)+B(x,k)+B(h,k)$, so the remainder after subtracting the proposed linear part $T(h,k):=B(h,y)+B(x,k)$ is exactly $B(h,k)$.

The map $T(h,k)=B(h,y)+B(x,k)$ is linear in $(h,k)$ and bounded: $\mathrm{\parallel }T(h,k)\mathrm{\parallel }\le C\mathrm{\parallel }h\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }+C\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }k\mathrm{\parallel }\le C(\mathrm{\parallel }y\mathrm{\parallel }+\mathrm{\parallel }x\mathrm{\parallel })\mathrm{\parallel }(h,k){\mathrm{\parallel }}_{\max }$ by [L1] and [L2].

For $(h,k)\mathrm{\ne }(0,0)$ the normalised remainder is $\mathrm{\parallel }B(h,k)\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }(h,k){\mathrm{\parallel }}_{\max }\le C\mathrm{\parallel }h\mathrm{\parallel }\mathrm{\parallel }k\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }(h,k){\mathrm{\parallel }}_{\max }\le C\mathrm{\parallel }(h,k){\mathrm{\parallel }}_{\max }$, which tends to $0$ as $(h,k)\to 0$ by [L1] and [L2]; hence $DB(x,y)=T$ by [L3].

For an associative algebra multiplication that is bounded bilinear, [step 2.1] with $B=m$ gives $Dm(a,b)(h,k)=m(h,b)+m(a,k)=hb+ak$, using bilinearity to write $m(h,b)=hb$ and $m(a,k)=ak$.

Assume $X=Y$. The diagonal map $d(x):=B(x,x)$ is then the well-typed composite of $x\mapsto (x,x)$ from $X$ to $X\times X$ with $B:X\times X\to Z$; the diagonal is bounded linear with derivative $h\mapsto (h,h)$, so the chain rule [L4] and [step 2.1] give $Dd(x)h=DB(x,x)(h,h)=B(h,x)+B(x,h)$.

Steps 2.1, 3.1 and 3.2 establish every displayed claim of the example.

Fix $y\in X$ and put $T_y(x):=y-g(x)$. Then $\mathrm{\parallel }{T}_y(x)-T_y(x^{\mathrm{\prime }})\mathrm{\parallel }=\mathrm{\parallel }g(x^{\mathrm{\prime }})-g(x)\mathrm{\parallel }\le q\mathrm{\parallel }x-x^{\mathrm{\prime }}\mathrm{\parallel }$ by [L1], so $T_y$ is a contraction of the nonempty complete metric space $X$; by [L2] it has exactly one fixed point, and $x=T_y(x)$ is equivalent to $F(x)=y$.

Consequently $F$ is bijective with $F^{-1}(y)$ the unique fixed point of $T_y$ for each $y$. If $x_i:=F^{-1}(y_i)$ for $i=1,2$, then $\mathrm{\parallel }{x}_1-x_2\mathrm{\parallel }=\mathrm{\parallel }(y_1-y_2)-(g(x_1)-g(x_2))\mathrm{\parallel }\le \mathrm{\parallel }{y}_1-y_2\mathrm{\parallel }+q\mathrm{\parallel }{x}_1-x_2\mathrm{\parallel }$, hence $\mathrm{\parallel }{F}^{-1}(y_1)-F^{-1}(y_2)\mathrm{\parallel }\le (1-q{)}^{-1}\mathrm{\parallel }{y}_1-y_2\mathrm{\parallel }$ because $1-q>0$.

Now assume $g$ is of class $C^k$ with $k\ge 1$; then $F$ is of class $C^k$ and $DF(x)=I_X+Dg(x)$ by [L5]. The derivative of $g$ satisfies $\mathrm{\parallel }Dg(x)\mathrm{\parallel }\le q$ by [L4], so $\mathrm{\parallel }-Dg(x)\mathrm{\parallel }\le q<1$ and [L3] makes $DF(x)=I_X-(-Dg(x))$ invertible with $\mathrm{\parallel }DF(x{)}^{-1}\mathrm{\parallel }\le (1-q{)}^{-1}$ at every $x$.

By the inverse function theorem [L5] applied at each $x$, and using that $F$ is a bijection by [step 2.1], the global inverse $F^{-1}$ agrees near each $y$ with the $C^k$ local inverse of $F$; being locally of class $C^k$, $F^{-1}$ is of class $C^k$. Hence $F$ is a global $C^k$ diffeomorphism.

Steps 2.1, 2.2 and 3.1 prove all the assertions of the example.

By [L1] the projection $p$ is bounded linear, so $Dp(x)=p$ for every $x$ by [L2]; it is surjective because $p(k+y)=y$ for every $y$, and its kernel is $\ker p=\{k+0:k\in K\}=K$ (Linear subspace of a vector space), which is complemented in $X$ by the given direct sum.

For every $y\in Y$ one has $p(k+y)=y$ for all $k\in K$, and conversely $p(x)=y$ forces $x=(x-y)+y$ with $x-y\in \ker p=K$; hence $p^{-1}(y)=K+y$, a translate of the subspace $K$.

Since [step 1.1] verifies the hypotheses of the regular value theorem at every point of every level set, that theorem gives that each $p^{-1}(y)$ is a split smooth submanifold of $X$ with $T_x{p}^{-1}(y)=\ker Dp(x)=K$ for all $x\in p^{-1}(y)$; by [step 2.1] this submanifold is the affine set $K+y$.

Every $y\in Y$ is therefore a regular value with the stated affine fibre and tangent space, which is the example's claim.

By [L1] the map $p$ is bounded linear with $\ker p=N$ finite dimensional, $\mathrm{ran}p=Y$ closed, and $\mathrm{coker}p=Y\mathrm{/}Y=\{0\}$ finite dimensional; hence $p$ is Fredholm at every point with index $\dim N-0=\dim N$ by [L3].

The map $p$ is smooth and $Dp(x)=p$ for every $x$ by [L2], so $p$ is a smooth Fredholm map of index $\dim N$ by [step 1.1].

For the reduction, take the splitting $X=\ker p\oplus Y$ and the range complement $C=\{0\}$; the normal form of [L3] reads $(u,v)\mapsto (u,g(u,v))$ with $g$ valued in the zero space, so $g\equiv 0$ and the obstruction space is trivial; the coordinates are those of the direct sum itself, and no nontrivial correction term is produced.

Thus $p$ is a smooth Fredholm map of index $\dim N$ whose local reduction has zero obstruction space, as claimed.

$c_0$ is closed in ${\mathrm{\ell }}^{\mathrm{\infty }}$ and is a Banach space for the sup norm by [F1]; ${\mathrm{\ell }}^{\mathrm{\infty }}$ is not second countable and hence is not a Banach manifold in the library's sense, while the identity chart still makes the split-chart condition meaningful for the pair $({\mathrm{\ell }}^{\mathrm{\infty }},c_0)$ by [L1]. [F1, L1] 1.2 Suppose for contradiction that $c_0$ is complemented in ${\mathrm{\ell }}^{\mathrm{\infty }}$: by [L7] fix a bounded linear projection $P:{\mathrm{\ell }}^{\mathrm{\infty }}\to c_0$ of ${\mathrm{\ell }}^{\mathrm{\infty }}$ onto $c_0$, so that $Px=x$ for every $x\in c_0$. [assume-contra, L7] 1.3 Fix an enumeration $(q_n)$ of the rationals; for every irrational $x$ and every $k\ge 1$ let $n_k(x)$ be the least index $n$ not already used among $n_1(x),\dots ,n_{k-1}(x)$ with $q_n\in (x-\frac{1}{k},x+\frac{1}{k})$, and put $A_x:=\{n_k(x):k\ge 1\}$. [L3, L4, construct] 2.1 Each $A_x$ is infinite. If $x\mathrm{\ne }y$, choose $K$ so large that the intervals $(x-\frac{1}{k},x+\frac{1}{k})$ and $(y-\frac{1}{\mathrm{\ell }},y+\frac{1}{\mathrm{\ell }})$ are disjoint whenever $k,\mathrm{\ell }\ge K$. Any common index of $A_x$ and $A_y$ must therefore occur among the first $K-1$ choices for at least one of $x,y$, a finite set; hence $A_x\cap A_y$ is finite. It follows that $x\mapsto A_x$ is injective and $\{A_x\}$ is uncountable by [L3]. [step 1.3, L3, algebra] 3.1 Let $u_x$ be the indicator sequence of $A_x$ and let $Q:={\mathrm{\ell }}^{\mathrm{\infty }}\mathrm{/}{c}_0$ carry the quotient norm; then $u_x\in {\mathrm{\ell }}^{\mathrm{\infty }}$ and $[u_x]\mathrm{\ne }0$ because $u_x\mathrm{\notin }{c}_0$. For distinct $x_1,\dots ,x_m$, remove the finite union of all pairwise intersections from their supports. The resulting indicators have disjoint supports and differ from the $u_{x_j}$ by finitely supported, hence $c_0$, sequences. Therefore, for scalars $c_1,\dots ,c_m$, the quotient norm of ${\sum }_j{c}_j[u_{x_j}]$ equals ${\max }_j\mathrm{\mid }{c}_j\mathrm{\mid }$: the disjoint representative gives the upper bound, and each remaining infinite support attains the corresponding coefficient infinitely often, giving the lower bound against every $c_0$ perturbation. [step 1.3, step 2.1, F1, L6, algebra] 4.1 For every $g\in Q^{\ast }$ and every real $r>0$ the set $\{x:\mathrm{\mid }g([u_x])\mathrm{\mid }\ge r\}$ is finite: for distinct points $x_1,\dots ,x_m$ in it choose unimodular scalars $c_j$ with $c_{j}g([u_{x_j}])=\mathrm{\mid }g([u_{x_j}])\mathrm{\mid }$, so that by [step 3.1] and [L8] one has $mr\le {\sum }_j\mathrm{\mid }g([u_{x_j}])\mathrm{\mid }=\mid g({\sum }_j{c}_j[u_{x_j}])\mid \le \mathrm{\parallel }g\mathrm{\parallel }\cdot \parallel {\sum }_j{c}_j[u_{x_j}]{\parallel }_Q=\mathrm{\parallel }g\mathrm{\parallel }$ and hence $m\le \mathrm{\parallel }g\mathrm{\parallel }\mathrm{/}r$. [step 3.1, L8, choose, algebra] 5.1 No countable family in $Q^{\ast }$ separates the points of $Q$: given $(g_n)$, the set of $x$ with $g_n([u_x])\mathrm{\ne }0$ for some $n$ is the countable union over the pairs $(n,k)$ of the finite sets of [step 4.1] with $r=\frac{1}{k}$, hence countable by [L5]; since $\{A_x\}$ is uncountable by [step 2.1], some $x$ lies outside it, and then the nonzero vector $[u_x]$ is annihilated by every $g_n$. [step 4.1, step 2.1, L5] 6.1 Let $P$ be the projection assumed in step 1.2 and put ${\psi }_n(x+c_0):=x_n-(Px{)}_n$. By [L2] each ${\psi }_n$ is a well-defined bounded linear functional on $Q$ — well-defined because $P$ fixes every element of $c_0$ — with $\mathrm{\parallel }{\psi }_n\mathrm{\parallel }\le 1+\mathrm{\parallel }P\mathrm{\parallel }$ by [L8], and if ${\psi }_n(x+c_0)=0$ for all $n$ then $(x-Px{)}_n=0$ for all $n$, so $x=Px\in c_0$. The countably many functionals ${\psi }_n$ would therefore be a countable separating family in $Q^{\ast }$, contradicting [step 5.1]. [step 1.2, step 5.1, L2, L6, L8] 7.1 This contradiction with [step 5.1] shows that no bounded linear projection of ${\mathrm{\ell }}^{\mathrm{\infty }}$ onto $c_0$ exists; by [L7] $c_0$ is not complemented in ${\mathrm{\ell }}^{\mathrm{\infty }}$, although it is closed there by [step 1.1], and consequently the identity chart admits no split-chart decomposition of $c_0$, as claimed.