Banach space norm induced by inner product; differentiability

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?

• 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? – kimchi lover Jun 2 '19 at 13:17
• $f:V\rightarrow W$ 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 $x\in V$. – 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.
• @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 – peek-a-boo Jun 2 '19 at 22:17