Chain sum product and composition rules for Banach derivatives

For claim 1, write $T:=aD{f}_1(x)+bD{f}_2(x)$, a bounded linear operator by [L4], and $r(h):=a{f}_1(x+h)+b{f}_2(x+h)-a{f}_1(x)-b{f}_2(x)-Th$, which equals $a{r}_1(h)+b{r}_2(h)$ with $r_j(h):=f_j(x+h)-f_j(x)-D{f}_j(x)h$. Then $\mathrm{\parallel }r(h)\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }h\mathrm{\parallel }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }{r}_1(h)\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }h\mathrm{\parallel }+\mathrm{\mid }b\mathrm{\mid }\mathrm{\parallel }{r}_2(h)\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }h\mathrm{\parallel }$ for $h\mathrm{\ne }0$, and both terms tend to $0$ by [L1]; hence $a{f}_1+b{f}_2$ is differentiable at $x$ with derivative $T$, which is claim 1.

For claim 2 define $L(h):=B(Df(x)h,g(x))+B(f(x),Dg(x)h)$ for $h\in X$. Since $Df(x)$, $Dg(x)$ are linear and $B$ is bilinear, $L$ is linear, and $\mathrm{\parallel }L(h)\mathrm{\parallel }\le C(\mathrm{\parallel }Df(x)\mathrm{\parallel }\mathrm{\parallel }g(x)\mathrm{\parallel }+\mathrm{\parallel }f(x)\mathrm{\parallel }\mathrm{\parallel }Dg(x)\mathrm{\parallel })\mathrm{\parallel }h\mathrm{\parallel }$ by [L3] and [L4], so $L$ is a bounded linear operator $X\to W$.

For claim 3 let $T:=Df(x)$ and $S:=Dg(f(x))$, write $y:=f(x)$, $R(k):=g(y+k)-g(y)-Sk$ for $k\in Y$ with $y+k\in W_0$, and $r(h):=f(x+h)-f(x)-Th$. Put $k(h):=f(x+h)-f(x)=Th+r(h)$; then $g(f(x+h))-g(f(x))-STh=Sr(h)+R(k(h))$ identically in $h$. Given a real $\eta >0$, apply [L1] for $g$ at $y$ with $\epsilon :=\eta \mathrm{/}(2(\mathrm{\parallel }T\mathrm{\parallel }+1))$ to get ${\delta }_1>0$ with $\mathrm{\parallel }R(k)\mathrm{\parallel }\le \frac{\eta }{2(\mathrm{\parallel }T\mathrm{\parallel }+1)}\mathrm{\parallel }k\mathrm{\parallel }$ for $\mathrm{\parallel }k\mathrm{\parallel }<{\delta }_1$, and apply [L1] for $f$ at $x$ with $\epsilon :=1$ and with $\epsilon :=\eta \mathrm{/}(2(1+\mathrm{\parallel }S\mathrm{\parallel }))$ to get a single ${\delta }_2>0$ such that for $\mathrm{\parallel }h\mathrm{\parallel }<{\delta }_2$ both $\mathrm{\parallel }k(h)\mathrm{\parallel }\le (\mathrm{\parallel }T\mathrm{\parallel }+1)\mathrm{\parallel }h\mathrm{\parallel }$ and $\mathrm{\parallel }r(h)\mathrm{\parallel }\le \frac{\eta }{2(1+\mathrm{\parallel }S\mathrm{\parallel })}\mathrm{\parallel }h\mathrm{\parallel }$ hold (take the smaller of the two thresholds). Then for $\mathrm{\parallel }h\mathrm{\parallel }<\min \{{\delta }_2,{\delta }_1\mathrm{/}(\mathrm{\parallel }T\mathrm{\parallel }+1)\}$ one has $\mathrm{\parallel }R(k(h))\mathrm{\parallel }\le \frac{\eta }{2}\mathrm{\parallel }h\mathrm{\parallel }$ and $\mathrm{\parallel }Sr(h)\mathrm{\parallel }\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }r(h)\mathrm{\parallel }\le \frac{\eta }{2}\mathrm{\parallel }h\mathrm{\parallel }$, so $\mathrm{\parallel }g(f(x+h))-g(f(x))-STh\mathrm{\parallel }\le \eta \mathrm{\parallel }h\mathrm{\parallel }$. Hence the bounded linear operator $ST$ of [L4] satisfies the remainder condition for $g\circ f$ at $x$, and by [L5] it is the derivative, which is claim 3.

For claim 2 put ${\mathrm{\Delta }}_f(h):=f(x+h)-f(x)$, ${\mathrm{\Delta }}_g(h):=g(x+h)-g(x)$, so that ${\mathrm{\Delta }}_f(h)=Df(x)h+r_f(h)$ and ${\mathrm{\Delta }}_g(h)=Dg(x)h+r_g(h)$ with remainders $r_f,r_g$ as in [L1]. By bilinearity, expanding $B(f(x)+{\mathrm{\Delta }}_f,g(x)+{\mathrm{\Delta }}_g)$ gives $B(f(x+h),g(x+h))-B(f(x),g(x))-L(h)=B(r_f(h),g(x))+B(f(x),r_g(h))+B({\mathrm{\Delta }}_f(h),{\mathrm{\Delta }}_g(h)),$ because $B({\mathrm{\Delta }}_f,g(x))=B(Df(x)h,g(x))+B(r_f,g(x))$ and likewise in the second variable, while the cross term is $B({\mathrm{\Delta }}_f,{\mathrm{\Delta }}_g)$.

For $\mathrm{\parallel }h\mathrm{\parallel }$ small, [L1] with $\epsilon =1$ gives $\mathrm{\parallel }{\mathrm{\Delta }}_f(h)\mathrm{\parallel }\le (\mathrm{\parallel }Df(x)\mathrm{\parallel }+1)\mathrm{\parallel }h\mathrm{\parallel }$ and $\mathrm{\parallel }{\mathrm{\Delta }}_g(h)\mathrm{\parallel }\le (\mathrm{\parallel }Dg(x)\mathrm{\parallel }+1)\mathrm{\parallel }h\mathrm{\parallel }$. Combining this with [step 2.1] and [L3], $\parallel B(f(x+h),g(x+h))-B(f(x),g(x))-L(h)\parallel \le C\mathrm{\parallel }{r}_f(h)\mathrm{\parallel }\mathrm{\parallel }g(x)\mathrm{\parallel }+C\mathrm{\parallel }f(x)\mathrm{\parallel }\mathrm{\parallel }{r}_g(h)\mathrm{\parallel }+C(\mathrm{\parallel }Df(x)\mathrm{\parallel }+1)(\mathrm{\parallel }Dg(x)\mathrm{\parallel }+1)\mathrm{\parallel }h{\mathrm{\parallel }}^2\mathrm{.}$ Dividing by $\mathrm{\parallel }h\mathrm{\parallel }$ for $h\mathrm{\ne }0$ and letting $h\to 0$, every term tends to $0$ by [L1], so the left-hand side is $o(\mathrm{\parallel }h\mathrm{\parallel })$ and the bounded linear operator $L$ of [step 1.2] satisfies the remainder condition for $B(f,g)$ at $x$; by [L5] it is the derivative, which is claim 2.

Claim 1 is [step 1.1], claim 2 is [step 3.1], and claim 3 is [step 1.3]; this is exactly the conjunction stated.