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?

  • $\begingroup$ 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

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.

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  • $\begingroup$ 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

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