Inverse function theorem for Banach spaces

(Affine changes of variable preserve the class.) Let $F$ be of class $C^k$ on an open set, let $\tau (x)=x_0+x$ be a translation, and let $L\in \mathcal{B}(Y,Z)$. Then $F\circ \tau$ is of class $C^k$ with $D^j(F\circ \tau )(x)=D^{j}F(\tau (x))$, and $L\circ F$ is of class $C^k$ with $D^j(L\circ F)(x)=L\circ D^{j}F(x)$, for every $j\le k$: by [L5] the first derivatives are $D(F\circ \tau )(x)=DF(\tau (x))$ and $D(L\circ F)(x)=L\circ DF(x)$, and the induction step differentiates these identities, the derivative of the translation being the identity and that of the bounded linear postcomposition $S\mapsto L\circ S$ being itself, with $\mathrm{\parallel }L\circ S\mathrm{\parallel }\le \mathrm{\parallel }L\mathrm{\parallel }\mathrm{\parallel }S\mathrm{\parallel }$ by [L6] ensuring continuity of the resulting expressions.

If $X=\{0\}$, then the isomorphism $A:X\to Y$ forces $Y=\{0\}$ and the theorem is immediate with $U_0=U=\{0\}$ and $V_0=Y$; hence assume $X\mathrm{\ne }\{0\}$, so $\mathrm{\parallel }{A}^{-1}\mathrm{\parallel }>0$. Since $Df$ is continuous at $a$ by [L1], choose $\rho >0$ such that $B(a,\rho )\subseteq U$ and $\mathrm{\parallel }Df(x)-A\mathrm{\parallel }<\frac{1}{2\mathrm{\parallel }{A}^{-1}\mathrm{\parallel }}(x\in B(a,\rho ))\mathrm{.}$ Put $r:=\rho \mathrm{/}3$. Then $\bar{B}(a,r)\subseteq B(a,2r)\subseteq B(a,\rho )\subseteq U$, and $\mathrm{\parallel }{A}^{-1}(Df(x)-A)\mathrm{\parallel }\le \frac{1}{2}$ on $B(a,2r)$. Thus $Df(x)$ is invertible there by [L4], with $\mathrm{\parallel }Df(x{)}^{-1}\mathrm{\parallel }\le 2\mathrm{\parallel }{A}^{-1}\mathrm{\parallel }$.

Put $\mathrm{\Phi }:=A^{-1}\circ f$ on $U$; by [L5], $D\mathrm{\Phi }(x)=A^{-1}\circ Df(x)$, so $\mathrm{\Phi }(a)=A^{-1}f(a)=:b$, $D\mathrm{\Phi }(a)=I_X$, and $\mathrm{\Phi }$ is of class $C^k$ by [step 1.1]. Put also $\mathrm{\Psi }(x):=x-\mathrm{\Phi }(x)$ on $B(a,2r)$; then $D\mathrm{\Psi }(x)=I_X-A^{-1}Df(x)=-A^{-1}(Df(x)-A)$, so $\mathrm{\parallel }D\mathrm{\Psi }(x)\mathrm{\parallel }\le \mathrm{\parallel }{A}^{-1}\mathrm{\parallel }\mathrm{\parallel }Df(x)-A\mathrm{\parallel }<\frac{1}{2}$ for every $x\in B(a,2r)$.

For $x,z\in \bar{B}(a,r)$ their segment lies in $\bar{B}(a,r)\subset B(a,2r)$, so [L2] applied to $\mathrm{\Psi }$ on the open convex set $B(a,2r)$ gives $\mathrm{\parallel }\mathrm{\Psi }(x)-\mathrm{\Psi }(z)\mathrm{\parallel }\le \frac{1}{2}\mathrm{\parallel }x-z\mathrm{\parallel }$. Since $\mathrm{\Phi }(x)-\mathrm{\Phi }(z)=(x-z)-(\mathrm{\Psi }(x)-\mathrm{\Psi }(z))$, also $\mathrm{\parallel }\mathrm{\Phi }(x)-\mathrm{\Phi }(z)\mathrm{\parallel }\ge \frac{1}{2}\mathrm{\parallel }x-z\mathrm{\parallel }$. In particular, $\mathrm{\Phi }$ is injective on $B(a,r)$.

Fix $y\in B(b,r\mathrm{/}4)$ and define $T_y(x):=y+x-\mathrm{\Phi }(x)=y+\mathrm{\Psi }(x)$ on $\bar{B}(a,r)$. For $x\in \bar{B}(a,r)$ one has $T_y(x)-a=(y-b)+\mathrm{\Psi }(x)-\mathrm{\Psi }(a)$ because $a-\mathrm{\Phi }(a)=\mathrm{\Psi }(a)$, so [step 3.1] yields $\mathrm{\parallel }{T}_y(x)-a\mathrm{\parallel }\le \mathrm{\parallel }y-b\mathrm{\parallel }+\frac{1}{2}\mathrm{\parallel }x-a\mathrm{\parallel }<\frac{r}{4}+\frac{r}{2}<r$; thus $T_y$ maps $\bar{B}(a,r)$ into $B(a,r)$, and it is a contraction with constant $\frac{1}{2}$ by [step 3.1]. By [L7] the closed ball is a nonempty complete metric space, so [L3] gives a unique fixed point $g(y)\in B(a,r)$ of $T_y$, and $T_y(x)=x$ is equivalent to $\mathrm{\Phi }(x)=y$; hence $y$ has exactly one preimage under $\mathrm{\Phi }$ in $B(a,r)$.

The set $W:=B(a,r)\cap {\mathrm{\Phi }}^{-1}(B(b,r\mathrm{/}4))$ is open and contains $a$, and $\mathrm{\Phi }{\mathrm{\mid }}_W:W\to B(b,r\mathrm{/}4)$ is a bijection with inverse $g$: it is injective by [step 3.1], and surjective by [step 4.1], which for each $y\in B(b,r\mathrm{/}4)$ produces $g(y)\in B(a,r)$ with $\mathrm{\Phi }(g(y))=y$. Moreover $g$ is Lipschitz with constant $2$: [step 3.1] gives $\mathrm{\parallel }g(y)-g(z)\mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\Phi }(g(y))-\mathrm{\Phi }(g(z))-(g(y)-g(z))\mathrm{\parallel }+\mathrm{\parallel }y-z\mathrm{\parallel }\le \frac{1}{2}\mathrm{\parallel }g(y)-g(z)\mathrm{\parallel }+\mathrm{\parallel }y-z\mathrm{\parallel }$.

For $y\in B(b,r\mathrm{/}4)$ put $x:=g(y)$; by [step 1.2] and [L4] the operator $D\mathrm{\Phi }(x)$ is invertible with $\mathrm{\parallel }D\mathrm{\Phi }(x{)}^{-1}\mathrm{\parallel }\le 2$. Let $h$ be small with $y+h\in B(b,r\mathrm{/}4)$ and put $k:=g(y+h)-g(y)$, so that $\mathrm{\parallel }k\mathrm{\parallel }\le 2\mathrm{\parallel }h\mathrm{\parallel }$ by [step 5.1] and $h=\mathrm{\Phi }(x+k)-\mathrm{\Phi }(x)=D\mathrm{\Phi }(x)k+r_{\mathrm{\Phi }}(k)$ with $\mathrm{\parallel }{r}_{\mathrm{\Phi }}(k)\mathrm{\parallel }=o(\mathrm{\parallel }k\mathrm{\parallel })$ by [L8]; applying $D\mathrm{\Phi }(x{)}^{-1}$ gives $k-D\mathrm{\Phi }(x{)}^{-1}h=-D\mathrm{\Phi }(x{)}^{-1}{r}_{\mathrm{\Phi }}(k)$, and $\mathrm{\parallel }D\mathrm{\Phi }(x{)}^{-1}{r}_{\mathrm{\Phi }}(k)\mathrm{\parallel }\le 2\mathrm{\parallel }{r}_{\mathrm{\Phi }}(k)\mathrm{\parallel }$, which is $o(\mathrm{\parallel }k\mathrm{\parallel })=o(\mathrm{\parallel }h\mathrm{\parallel })$; hence $g$ is differentiable at $y$ with $Dg(y)=D\mathrm{\Phi }(x{)}^{-1}=Df(g(y){)}^{-1}A$.

The derivative formula of [step 6.1] is continuous in $y$: the map $y\mapsto D\mathrm{\Phi }(g(y))$ is continuous because $D\mathrm{\Phi }$ is continuous by [L1] and [step 2.1] and $g$ is continuous by [step 5.1]; inversion is continuous at each invertible operator, since for $\mathrm{\parallel }{A}^{-1}E\mathrm{\parallel }\le \frac{1}{2}$ the bound $\mathrm{\parallel }(A+E{)}^{-1}-A^{-1}\mathrm{\parallel }\le \mathrm{\parallel }(A+E{)}^{-1}\mathrm{\parallel }\mathrm{\parallel }E\mathrm{\parallel }\mathrm{\parallel }{A}^{-1}\mathrm{\parallel }\le 2\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}^2\mathrm{\parallel }E\mathrm{\parallel }$ from [L4] and [L6] tends to $0$ with $\mathrm{\parallel }E\mathrm{\parallel }$. Hence $g$ is of class $C^1$.

(Higher regularity.) Inversion $\mathrm{i}\mathrm{n}\mathrm{v}(A):=A^{-1}$ is of class $C^{\mathrm{\infty }}$ on the open set of invertible operators in $\mathcal{B}(X)$: for $\mathrm{\parallel }{A}^{-1}H\mathrm{\parallel }<1$, [L4] gives $(A+H{)}^{-1}=A^{-1}-A^{-1}H{A}^{-1}+R(H)$ with $\mathrm{\parallel }R(H)\mathrm{\parallel }\le \frac{\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}^3\mathrm{\parallel }H{\mathrm{\parallel }}^2}{1-\mathrm{\parallel }{A}^{-1}H\mathrm{\parallel }}$, whence $D\mathrm{i}\mathrm{n}\mathrm{v}(A)H=-A^{-1}H{A}^{-1}$; that formula is continuous in $A$ by [L4] and [L6], and iterating the expansion differentiates it again, so $\mathrm{i}\mathrm{n}\mathrm{v}$ is $C^r$ for every $r$. Now let $k\ge 2$ and let $\mathrm{\Phi }$ be of class $C^k$; then $D\mathrm{\Phi }$ is of class $C^{k-1}$ by [L1] and $Dg=\mathrm{i}\mathrm{n}\mathrm{v}\circ D\mathrm{\Phi }\circ g$ by [step 6.1] and [L5]. If $g$ is of class $C^{j-1}$ for some $j$ with $2\le j\le k$, then $Dg$, a composite of $C^{j-1}$ maps, is of class $C^{j-1}$ and hence $g$ is of class $C^j$; the case $j=1$ is [step 7.1], so induction gives $g$ of class $C^k$.

Transfer to $f$: by [step 5.1] the set $U_0:=W$ is an open neighbourhood of $a$ contained in $U$, and $V_0:=f[U_0]=A[\mathrm{\Phi }[W]]=A[B(b,r\mathrm{/}4)]$ is an open neighbourhood of $f(a)$; the restriction $f{\mathrm{\mid }}_{U_0}$ is a bijection onto $V_0$ with inverse $F(y):=g(A^{-1}y)$, which is of class $C^k$ by [step 1.1] and [step 8.1] as a composite of the bounded linear map $A^{-1}$ and the $C^k$ map $g$.

For $x\in U_0$ the map $F\circ f$ equals the identity near $x$, so the chain rule [L5] differentiates it to $DF(f(x))Df(x)=I_X$; multiplying on the right by $Df(x{)}^{-1}$ gives $DF(f(x))=Df(x{)}^{-1}$, which is the displayed derivative formula for the statement's inverse $g$, here the map $F$ of [step 9.1].