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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.

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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 $$ Df(x,y)(h,k) = f(x,k) + f(h,y) $$

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

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