Fréchet derivative of $x\mapsto{\rm D}A(x)\circ B(x)$ with $A:H\to H$ and $B:H\to\mathfrak L(U,H)$

Question

I guess the answer of this question is easy, but I just can't figure it out at the moment:

Let

• $U,H$ be Hilbert spaces
• $A:H\to H$ be twice continuously Fréchet differentiable
• $B:H\to\mathfrak L(U,H)$ be continuously Fréchet differentiable

Now, let $$\Phi(x):={\rm D}A(x)\circ B(x)\;\;\;\text{for }x\in H\;.$$ How can we calculate ${\rm D}\Phi$?

Answer 1

Form your preconditions, $DA(x)$ is a linear and continous map from $H$ to $H$. Using the chain rule you otain

$$D\Phi(x)y={\rm D (DA(x)B(x)y)}\circ D(B(x)y)$$

As $DA(x)$ is linear this simplifies to

$$D\Phi(x)y=DA(x)\circ D(B(x)y)$$

If you don't know anything about $B$, that should be the result (You know that B is coniuously Frechet differentiable). If $B(x)$ is linear and continous this simplifies further and you achieve m.s' result.