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.


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

  • $\begingroup$ Why do the inequalities follow? $\endgroup$ – Polp Feb 29 '16 at 23:02
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.