Let $E,F$ be Banach spaces. If $f\colon [0,1]\times E\to F$ is continuously differentiable, can we say something about the differentiability of $$g\colon E\to C^1([0,1],F),\quad y\mapsto f(\cdot,y)?$$ Is $(\partial g(y)h)(t) = \partial_2f(t,y)h$ for $h\in E$?
I am a bit confused. Can someone clear this situation up? Thanks a lot.
The question can be answered positively.
Consider the superposition operator $f^\sharp$ induced by $f$: $$f^\sharp\colon E^{[0,1]}\to F^{[0,1]},\quad u\mapsto f(\cdot,u(\cdot)).$$
Because $f$ is continuously differentiable it is $f^\sharp\in C^1(C([0,1],E),C([0,1],F))$ and $$\left(\partial f^{\sharp}(u)v\right)(t) = \partial_2f(t,u(t))v(t),\quad t\in[0,1],\quad u,v\in C([0,1],E),$$ as is shown in e.g. “Analysis II” by H.Amman & J. Escher.
The answer follows from identifying $E = C_{\operatorname{const}}([0,1],E)$ and noting that $[y\mapsto f(\cdot,y)] = f^{\sharp}|_{C_{\operatorname{const}}([0,1],E)}.$