# Frechet derivative for bilinear map

Let $\mathbb{X}, \mathbb{Y}$ and $\mathbb{Z}$ be normed spaces and let $f$ be a bounded bilinear map. Show that $f$ is Frechet differentiable at every $(x,y) \in \mathbb{X} \times \mathbb{Y}$ and find its Frechet derivative. (View f as a map $\mathbb{X} \times \mathbb{Y} \rightarrow \mathbb{Z}$).

What I have tried: I think the derivative is the function itself but I'm not sure how to set it out formally. I know how to do it for single normed spaces but I am confused with the bilinear map stuff.

## 1 Answer

You have to consider $X \times Y$ as a single normed space.

Let $(x,y), (h,k) \in X \times Y$. We have \begin{align*} f(x+h, y+k) &= f(x,y) + f(x,k) + f(h,y) + f(h,k)\\ \end{align*} As $f$ is bounded, $$\|f(h,k)\| \le c\|h\|\|k\| \le c\|(h,k)\|^2 = o(\|(h,k)\|)$$ Now, $$(h,k) \mapsto f(x,k) + f(h,y)$$ is a bounded linear map $X \times Y \to Z$, hence we have $$Df(x,y)(h,k) = f(x,k) + f(h,y)$$

• Why do the inequalities follow? – Polp Feb 29 '16 at 23:02
• A bilinear map is bounded, iff $$\sup\{ \|f(h,k)\| : \|h\| = \|k\| = 1 \}$$ is finite .. as $f$ is bounded we have the inequality with $c$ the sup above. – martini Mar 1 '16 at 13:47
• What about the second inequality? – Polp Mar 1 '16 at 13:50
• That holds by the very definition of the norm on $X \times Y$, we have - as one of many equivalent norms $$\|(h,k)\| := \|h\| + \|k\|$$ – martini Mar 1 '16 at 13:57
• Martini, could you explain again why $||h||||k|| = ||(h,k)||^2$? – jpugliese Feb 25 '18 at 21:36