0
$\begingroup$

Let $\left( X,\left\| \cdot \right\| \right)$ be a Banach space where the norm $\left\| \cdot \right\|$ is induced by an inner product $\langle \cdot ,\cdot \rangle$.
Let $f:X\times X\rightarrow X$, $f\left( x,y\right) =\langle y,x\rangle x$.

Show that f is differentiable in every point and calculate the total derivative.

I tried to show that all directional derivatives exist and are continuous but that didn‘t work. Is there a better way?

$\endgroup$
  • $\begingroup$ What's your definition of differentiable? (There are more than 1 in Banach spaces, or even (I think) in Hilbert spaces.) And "differentiable" where? At the origin? Everywhere? $\endgroup$ – kimchi lover Jun 2 '19 at 13:17
  • $\begingroup$ $f:V\rightarrow W$ is called differentiable in $a\in V$, if a linear mapping $L:V\rightarrow W$ and a function $R:V\rightarrow W$ exist, so that $f\left( x\right) =f\left( a\right) +L\left( x-a\right) +R\left( x\right)$ and $\lim _{v\rightarrow 0}\dfrac {R\left( x\right) }{\left\| x-a\right\| }=0$ for all $x\in V$. $\endgroup$ – Moe1234 Jun 2 '19 at 13:30
0
$\begingroup$

For Frechet differentiability, start by computing $$ f(x+h,y+k)-f(x,y)=\langle y+k,x+h\rangle(x+h)-\langle y,x\rangle x=\dots $$ and show it is "linear in $(h,k)$" + smaller.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I haven‘t learned about Frechet differentiability yet. Do you know of a way to show this with the definition of differentiability in the second comment under the post? $\endgroup$ – Moe1234 Jun 2 '19 at 13:38
  • $\begingroup$ @Moe1234 I think you meant $\lim_{x \to a}$ not $\lim_{v \to 0}$. In this case, what you have written there is equivalent to Frechet differentiability $\endgroup$ – peek-a-boo Jun 2 '19 at 22:17
  • $\begingroup$ Yes i meant x->a. Thanks $\endgroup$ – Moe1234 Jun 3 '19 at 13:45

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.