# If $G\in C^2(H,\mathfrak L(U,H))$, what's the Fréchet derivative of $\Phi(x)(u,v):=({\rm D}G(x)(G(x)u))v$?

Let

• $U,H$ be separable $\mathbb R$-Hilbert spaces
• $\mathfrak L(A,B)$ denote the space of bounded linear operators between normed $\mathbb R$-vector spaces $A,B$
• $G:H\to\mathfrak L(U,H)$ be twice continuously Fréchet differentiable and ${\rm D}G:H\to\mathfrak L(H,\mathfrak L(U,H))$ and ${\rm D}^2G:H\to\mathfrak L(H,\mathfrak L(H,\mathfrak L(U,H)))$ denote the first and second Fréchet derivatives of $G$, respectively

How can we calculate the Fréchet derivative of $\Phi:H\to\mathfrak L^{(2)}(U,H)$ with $$\Phi(x)(u,v):=({\rm D}G(x)(G(x)u))v\;\;\;\text{for }x\in H\text{ and }u,v\in U\;,$$ where $\mathfrak L^{(2)}(U,H)$ denotes the space of bounded bilinear operators from $U\times U$ to $H$?

The Fréchet derivative is $$\big(D\Phi(x)[h]\big)(u,v) = \big(D^2G(x)[h]\big)\big[G(x)[u]\big][v] + DG(x)\big[\big(DG(x)[h]\big)[u]\big][v],$$ where $[\cdot]$ implies a linear input, and $()$ is used for grouping and input which might not be linear.
Consider the following calculation \begin{aligned} &\frac{\|\Phi(x+h) - \Phi(x) - D\Phi(x)[h]\|_{L(U\times U,H)}}{\|h\|_H} \\ &\qquad= \sup_{\|(u,v)\|_{U\times U}=1}\frac{\|\Phi(x+h)(u,v) - \Phi(x)(u,v) - \big(D\Phi(x)[h]\big)(u,v)\|_H}{\|h\|_H}. \end{aligned}\tag{\ast} We will consider the $H$-norm expression. \begin{aligned} \|\Phi&(x+h)(u,v) - \Phi(x)(u,v) - \big(D\Phi(x)[h]\big)(u,v)\|_H \\ &=\left\|\big(DG(x+h)\big[G(x+h)[u]\big]\big)[v] - DG(x)\big[G(x)[u]\big][v] \right.\\ &\qquad\qquad \left.- \big(D^2G(x)[h]\big)\big[G(x)[u]\big][v] - DG(x)\big[\big(DG(x)[h]\big)[u]\big][v]\right\|_H \\ &\leq\left\|\big(DG(x+h)\big[G(x+h)[u]\big]\big) - DG(x)\big[G(x)[u]\big] \right.\\ &\qquad\qquad \left.- \big(D^2G(x)[h]\big)\big[G(x)[u]\big] - DG(x)\big[\big(DG(x)[h]\big)[u]\big]\right\|_{L(U,H)}\|v\|_U. \end{aligned} From here we drop the $v$ factor and simply consider the $L(U,H)$-norm factor. Here we add and subtract the terms $DG(x)\big[G(x+h)[u]\big] + \big(D^2G(x)[h]\big)\big[G(x+h)[u]\big]$, then collecting we get \begin{aligned} &\left\|\big(DG(x+h)-DG(x)-D^2G(x)[h]\big)\big[G(x+h)[u]\big] \right.\\ &\qquad\qquad+ DG(x)\big[\big(G(x+h) - G(x) - DG(x)[h]\big)[u]\big] \\ &\qquad\qquad\qquad\qquad \left.+ \big(D^2G(x)[h]\big)\big[\big(G(x+h)-G(x)\big)[u]\big]\right\|_{L(U,H)} \\ &\leq \left\|\big(DG(x+h)-DG(x)-D^2G(x)[h]\big)\big[G(x+h)[u]\big] \right\|_{L(U,H)} \\ &\qquad\qquad+ \left\|DG(x)\big[\big(G(x+h) - G(x) - DG(x)[h]\big)[u]\big]\right\|_{L(U,H)} \\ &\qquad\qquad\qquad\qquad + \left\| \big(D^2G(x)[h]\big)\big[\big(G(x+h)-G(x)\big)[u]\big]\right\|_{L(U,H)} \\ &\leq \left\|DG(x+h)-DG(x)-D^2G(x)[h]\right\|_{L(H,L(U,H))}\|G(x+h)[u]\|_H \\ &\qquad\qquad+ \left\|DG(x)\right\|_{L(H,L(U,H))}\|\big(G(x+h) - G(x) - DG(x)[h]\big)[u]\|_H \\ &\qquad\qquad\qquad\qquad + \left\|D^2G(x)[h]\right\|_{L(H,L(U,H))}\|\big(G(x+h)-G(x)\big)[u]\|_H. \end{aligned} Going back to the initial expression ($\ast$) we get \begin{aligned} &\sup_{\|(u,v)\|_{U\times U}=1}\frac{\|\Phi(x+h)(u,v) - \Phi(x)(u,v) - \big(D\Phi(x)[h]\big)(u,v)\|_H}{\|h\|_H} \\ &\qquad \leq \sup_{\|(u,v)\|_{U\times U}=1}\left(\frac{\left\|DG(x+h)-DG(x)-D^2G(x)[h]\right\|_{L(H,L(U,H))}}{\|h\|_H}\right. \|G(x+h)[u]\|_H \\ &\qquad\qquad\qquad\qquad + \left\|DG(x)\right\|_{L(H,L(U,H))}\frac{\|\big(G(x+h) - G(x) - DG(x)[h]\big)[u]\|_H}{\|h\|_H} \\ &\qquad\qquad\qquad\qquad + \left.\frac{\left\|D^2G(x)[h]\right\|_{L(H,L(U,H))}}{\|h\|_H}\|\big(G(x+h)-G(x)\big)[u]\|_H\right)\|v\|_U. \\ &\qquad \leq \left(\frac{\left\|DG(x+h)-DG(x)-D^2G(x)[h]\right\|_{L(H,L(U,H))}}{\|h\|_H}\right. \|G(x+h)\|_{L(U,H)} \\ &\qquad + \left\|DG(x)\right\|_{L(H,L(U,H))}\frac{\|G(x+h) - G(x) - DG(x)[h]\|_{L(U,H)}}{\|h\|_H} \\ &\qquad + \left.\frac{\left\|D^2G(x)\right\|_{L(H,L(H,L(U,H)))}\|h\|_H}{\|h\|_H}\|G(x+h)-G(x)\|_{L(U,H)}\right)\sup_{\|(u,v)\|_{U\times U}=1}\|u\|_U\|v\|_U. \end{aligned} As $\|h\|_H \to 0$ the first and second term goes to zero by the Fréchet differentiability of $G$ and $DG$. The last term the factors $\|h\|_H$ cancel, but by continuity of $G$ it goes to zero as well. And thus we are done.