Prove that $\Phi$ is differentiable on $E$ and find $\Phi'(x)$

Question

Good day all, here is a problem on differentiabilty on $\Bbb{R}^n$ that I find difficult.

Let $E$ be a normed-vector space over $\Bbb{R}$ such that $\dim E<\infty$ and $F$ also, be a normed-vector space. Let $u,v\in L(E)$ and $f:E\times E\to F$ be bilinear. Here, $L(E)$ is the space of continuous linear maps from $E$ to $E$. Let $$\Phi:E\to F$$ $$x\mapsto \Phi(x)=f(u(x),v(x)).$$

I want to prove that $\Phi$ is differentiable on $E$ and as well, want to compute $\Phi'(x).$ I'm thinking that decomposing the function into composition of functions, will enable me do this job but I don't know how to go about it. Please, could anyone put me through? I have just been introduced to differentiability on $\Bbb{R}^n$. Thanks for your time and help!

Answer 1

Since $f$ is bilinear, $f$ is differentiable. Since $u$ and $v$ are differentiable, the function$$\begin{array}{ccc}E\times E&\longrightarrow&E\times E\\(x,y)&\mapsto&\bigl(u(x),v(y)\bigr)\end{array}$$is differentiable. Finally, the function$$\begin{array}{ccc}E&\longrightarrow&E\times E\\x&\mapsto&(x,x)\end{array}$$is differentiable too. Composing them, you get your function.

Answer 2

Let $x,h \in E$ then \begin{align} \Phi(x+h)=f(u(x+h),v(x+h))&= f(u(x)+u(h),v(x)+v(h))\\ &=f(u(x),v(x)+v(h))+f(u(h),v(x)+v(h))\\ &=f(u(x),v(x))+f(u(x),v(h)) +f(u(h),v(x))+f(u(h),v(h)) \end{align}

So $\Phi$ is differentiable at any point $x \in E$ and $$d\Phi_x(h)=f(u(x),v(h)) +f(u(h),v(x))$$ which mean that $$\Phi'(x)=f(u(x),v'(x)) +f(u'(x),v(x))$$