If I have a map $K\times K\rightarrow K$, which maps $(x,y)$ to the product $xy$.
How would i prove by direct caculation that the map is Frechet Differentiable without using the product rule.
I've been trying for a while but haven't really got anywhere. Any help would be greatly appreciated, thanks
If $f(x,y)=xy$ we have that $$f(x+h,y+h)= (x+h)(y+h)=xy+(x+y)h+h^2 \ .$$ Thus $$f(x+h,y+h)=f(x,y)+ D_{(x,y)}(h)+ o(h)$$ where $D_{(x,y)}(h)=(x+y)\cdot h$. Meaning that $f(x,y)$ is Frechet differentiable with differential $D_{(x,y)}$.
Calculating the Frechet derivate is sometimes very complicated (but obviously not in this case). For this reason it is good to know the following lemma:
Let $X$ be a banach space, $U\subset X$ open and $I\colon U \to \mathbf R$ Gateaux differentiable. Assume furter that $I_G'$ is continuous in $u\in U$. Then we have $$I_G'(u) = I_F'(u)$$ where $I_G'$ and $I_F'$ denotes the Gateaux derivate repectively Frechet derivate.
Set $I(x,y)=xy$. We have \begin{align*} \lim_{t\to 0} \frac{I(u_1+tv_1, u_2 +tv_2) - I(u_1,u_2)}{t} & = \lim_{t\to 0} \frac{(u_1 +tv_1)(u_2 +tv_2) - u_1u_2}{t} \\ &= \lim_{t\to 0} v_1u_2 +u_1v_2 +tv_1v_2 \\ &= v_1u_2 +u_1v_2 =: \langle A_u,v\rangle. \end{align*} Hence we have $$A_u=I_G'(u) = I_F'(u).$$