Local Banach implicit-function supplier

Uniform contraction bound. Define $T_{ϵ}(x):=x-L^{-1}F(x,ϵ)$ wherever $(x,ϵ)\in U$. Then $D_x{T}_0(0)=I-L^{-1}L=0$. By continuity of $D_{x}F$, choose $\rho >0$, ${ϵ}_1>0$, and $q\in (0,1)$ such that the closed product ${\bar{B}}_X(0,\rho )\times [-{ϵ}_1,{ϵ}_1]$ lies in $U$ and $\mathrm{\parallel }{D}_x{T}_{ϵ}(x)\mathrm{\parallel }\le q$ there. For fixed $ϵ$ and $x,y$ in the ball, the curve $s\mapsto T_{ϵ}(y+s(x-y))$ is continuous and differentiable with derivative norm at most $q\mathrm{\parallel }x-y\mathrm{\parallel }$; [F5] therefore gives $\mathrm{\parallel }{T}_{ϵ}(x)-T_{ϵ}(y)\mathrm{\parallel }\le q\mathrm{\parallel }x-y\mathrm{\parallel }$.

Fixed points and local uniqueness. Since $T_0(0)=0$, continuity of $F$ allows ${ϵ}_0\in (0,{ϵ}_1]$ such that $\mathrm{\parallel }{T}_{ϵ}(0)\mathrm{\parallel }<(1-q)\rho$ for $\mathrm{\mid }ϵ\mathrm{\mid }<{ϵ}_0$. Step 1.1 then gives $\mathrm{\parallel }{T}_{ϵ}(x)\mathrm{\parallel }\le q\rho +\mathrm{\parallel }{T}_{ϵ}(0)\mathrm{\parallel }<\rho$ whenever $\mathrm{\parallel }x\mathrm{\parallel }\le \rho$, so $T_{ϵ}$ maps the nonempty closed ball ${\bar{B}}_X(0,\rho )$ into itself and its fixed point lies in the open ball. It is complete by [F1] and [F6], and is a contraction by step 1.1. By [F3] it has exactly one fixed point $x(ϵ)$, and $T_{ϵ}(x)=x$ is equivalent to $F(x,ϵ)=0$ because $L^{-1}$ is injective. Thus the zero set in the product neighborhood is the graph of $x$, and $x(0)=0$.

Invertibility of nearby partial derivatives. Let $B(X)$ be the bounded linear operators on $X$, with the norm of [F7]. It is complete in the operator norm: an operator-norm Cauchy sequence converges pointwise to a linear map, the uniform norm bounds make that limit bounded, and the same Cauchy estimate gives convergence in operator norm. Shrinking the product neighborhood if necessary, continuity of $D_{x}F$ gives $\mathrm{\parallel }{L}^{-1}{D}_{x}F(x,ϵ)-I\mathrm{\parallel }\le q<1$. Put $E=L^{-1}{D}_{x}F-I$. The series ${\sum }_{j=0}^{\mathrm{\infty }}(-E{)}^j$ converges in $B(X)$ because ${\sum }_j\mathrm{\parallel }E{\mathrm{\parallel }}^j<\mathrm{\infty }$, and its finite partial sums multiplied on either side by $I+E$ converge to $I$; thus $D_{x}F(x,ϵ)$ is invertible. The inverse is continuous, since $A^{-1}-B^{-1}=A^{-1}(B-A)B^{-1}$ for invertible nearby operators with uniformly bounded inverses.

Continuity and a local Lipschitz bound. For $ϵ,\eta$ in a smaller interval, the fixed-point equations and step 1.1 yield $\mathrm{\parallel }x(ϵ)-x(\eta )\mathrm{\parallel }\le (1-q{)}^{-1}\mathrm{\parallel }{T}_{ϵ}(x(\eta ))-T_{\eta }(x(\eta ))\mathrm{\parallel }$. The derivative $D_{ϵ}F$ is bounded on a sufficiently small neighborhood of $(0,0)$, since it is continuous there; applying [F5] to the parameter segment bounds the right-hand side by $C\mathrm{\mid }ϵ-\eta \mathrm{\mid }$. Hence $x$ is locally Lipschitz, in particular continuous.

Differentiation of the branch. Write $\delta x=x(ϵ+h)-x(ϵ)$. Step 3.1 gives $\mathrm{\parallel }\delta x\mathrm{\parallel }=O(\mathrm{\mid }h\mathrm{\mid })$. Fréchet differentiability at $(x(ϵ),ϵ)$ yields $0=D_{x}F(x(ϵ),ϵ)\delta x+D_{ϵ}F(x(ϵ),ϵ)h+o(\mathrm{\parallel }\delta x\mathrm{\parallel }+\mathrm{\mid }h\mathrm{\mid })$. Divide by $h\mathrm{\ne }0$; the remainder divided by $\mathrm{\mid }h\mathrm{\mid }$ tends to zero because $\mathrm{\parallel }\delta x\mathrm{\parallel }=O(\mathrm{\mid }h\mathrm{\mid })$. Multiplying by the bounded inverse from step 2.2 gives $\delta x\mathrm{/}h\to -D_{x}F(x(ϵ),ϵ{)}^{-1}{D}_{ϵ}F(x(ϵ),ϵ)$. This proves differentiability and the displayed formula. The formula is continuous in $ϵ$ because $x$, $DF$, and the nearby operator inverse are continuous. Therefore $x$ is $C^1$ on a possibly smaller interval, which proves the statement.