Problem: Let $X$ and $Y$ be two normed linear spaces. A function $ϕ : X×Y → \mathbb{R}$ is bilinear if $f(·, y) ∈ L(X, \mathbb{R})$ and $f(x, ·) ∈ L(Y, \mathbb{R})$ for each $(x, y) ∈ X × Y$ . Show that any continuous bilinear function ϕ is Fréechet differentiable with $D_{ϕ,(x,y)}$ $(z, w)$ = $ϕ(z, y) + ϕ(x, w)$
I have attempted a proof but I require help on the last step.
Proof:
Let $\phi: X$ x Y$ \rightarrow \mathbb{R}$ be bilinear and continuous. Fix $(x,y) \in int(X $x$ Y)$. We have that $\phi$ is Fréchet Differentiable if $\exists D_{\phi,(x,y)}$ such that
$$\lim_{v\to 0} (1/||v||) * [\phi(x+v) - \phi(x) - D_{\phi,(x,y)}(v)] = 0 $$
for some $v \in V$, V a neighborhood of 0 in $X x Y$ such that $x+v \in int(X x Y), \forall v \in V$.
Take x = $(x_1,x_2)$ and v $ = (v_1, v_2)$
By bilinearity, $\phi(x + v)$ = $\phi(x_1, x_2) + \phi(v_1, x_2) + \phi(x, v_2) + \phi(v_1, v_2)$
We then have that
$$\lim_{v\to 0} (1/||v||) * [\phi(v_1, x_2) + \phi(x,v_2) + \phi(v_1, v_2) - D_{\phi,(x,y)}(v)] = 0 $$
A potential candidate for the existence of a Fréchet derivative such that this limit tends to zero is $D_{\phi,(x,y)}(v) = \phi(v_1, y) + \phi(x, v_2)$. If I prove that
$$\lim_{v\to 0} (1/||v||) * \phi(v_1, v_2) = 0 \space (2)$$
Then it means the Fréchet derivative exists and is equal to the one I proposed.
I have stopped here because I couldn't find a way to use continuity to get (2). Anyone have any idea of a norm I could choose or how I could use that? Is everything ok to that point?
Thanks