Proving the product rule for Fréchet Derivative.

Question

Let $X$ be a normed vector space, $U\subset X$, and $F,G:U\rightarrow \mathbb{R}$ differentiable at $x\in U$. Show that the map $F\cdot G:U\rightarrow \mathbb{R}$, $F\cdot G(x)=F(x)G(x)$ is also differentiable at $x\in U$ and that $$ D(F\cdot G)_{|_x}=F(x)DG_{|_x}+G(x)DF_{|_x} $$ AKA Prove the product rule for the Fréchet Derivative.

---

To be Fréchet differentiable means the following: Let $X,Y$ be normed vector spaces, U open in X, and $F:U\rightarrow Y$. Let $x,h\in U$ and let $T:X\rightarrow Y$ be a linear map. Then the limit $$ \underset{h\rightarrow 0}{\lim}\frac{||F(x+h)-F(x)-Th||_Y}{||h||_X}=0$$ exists. We denote $T$ as $DF_{|_x}$.

---

Here is my attempt at a proof.

$Proof.$

Let $$ S=F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h $$ By some algebraic manipulations we have, $$ =F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h + F(x+h)G(x)-F(x+h)G(x)$$ $$ =F(x+h)\big[G(x+h)-G(x)-DG_{|_x}h\big] + G(x)\big[F(x+h)-F(x)-DF_{|_x}\big] -F(x)DG_{|_x}h - G(x)DF_{|_x}h +F(x+h)DG_{|_x}h+G(x)DF_{|_x}h $$ $$ =F(x+h)\big[G(x+h)-G(x)-DG_{|_x}h\big] + G(x)\big[F(x+h)-F(x)-DF_{|_x}\big] + DG_{|_x}h\big[F(x+h)-F(x)\big] $$ Then since F and G are Fréchet differentiable at x, we have $$ \underset{h\rightarrow 0}{\lim}\frac{|S|}{||h||_X} $$ $$ =F(x+h)(0)+G(x)(0)+\underset{h\rightarrow 0}{\lim}\frac{|DG_{|_x}||h||F(x+h)-F(x)|}{||h||_X} $$

---

At the end there, I'm pretty sure that the $h$'s do not cancel since it is possible that $X$ is infinite dimensional and, therefore, the norms cannot be said to be equivalent. If it was finite dimensional, I would have argued by continuity of $F$ that $F(x+h)-F(x)\rightarrow 0$ as $h\rightarrow 0$.

Does anyone know how to proceed? Or a different approach to the problem

Answer 1

I'd prove using the following general rules of differentiation:

1. Chain rule $(g \circ f)'(x) = g'(f(x)) \circ f'(x)$;

2. Leibniz rule $B'(x,y) \cdot (h, k) = B(x, k) + B(h, y)$ for continuous bilinear functions.

3. Derivate of product spaces $(f, g)'(x) = (f'(x), g'(x)).$

Then, you want to differentiate $\psi(x) = F(x) G(x) = (B \circ (F,G)) (x),$ where $B$ is the bilinear function $(x, y) \mapsto xy.$

Answer 2

This might be a little bit late:

First, your computation is 100% right. However, you should note that $(DG)_{x}$ applied on $f$ is a bounded linear operator regardless of dimension of your space.

How so? Recall the equivalent definition of a bouned linear operator T, it requires:

$sup_{||f||=1} ||Tf||_R \lt \infty$, this is automatically satisfied by the definition of derivative at a point because it has to be finite.

However, for general linear operator $Df = f'$ maps $C^1[0,1]$ to $C^0[0,1]$， it is obviously not bounded.