Differentiation calculation

Question

$L(E)$ espace fonction continuous and linear $$\begin{array}{llll} \psi:& L(E)\times E&\longrightarrow& E\\ &(u,x)&\longrightarrow &u(x) \end{array}$$ proved the application $\psi$ differentiable and calculer this differential ?

Answer 1

A function $\psi: V \to R$ is differentiable at $p \in D$ when $$\lim_{v \to 0} \psi(p+v) - \psi(p) = L(p)v + r(v)$$ and $\frac{r(v)}{\|v\|}\to 0$ as $v \to 0$. (Here $V$ and $R$ are normed vector spaces)

To check for the existence of a derivative you take a general $v$ and try to guess $L(x): V \to R$.

In our case take $ p = (u,x)$ and $v= (d,h)$ (as in user251257 )

$$ \psi(p+v) - \psi(p) = \psi(u+d, x+h) - \psi(u,x) = (u+d)(x+h) - u(x) = u(x+h) + d(x+h) - u(x) = u(x)+ u(h) + d(x) + d(h) - u(x) = u(h) + d(x) + d(h) $$

Since $\|(d,h)\| \geq \frac{|h|}{2}$ it is reasonable to guess that the product $|d(h)| \leq \|d\||h|$ is an error term, so take $L(u,x)(d,h):= u(h)+ d(x)$ (note this is a linear transformation). Moreover

$$ \psi(u+d, x+h) - \psi(u,x) -L(u,x)(d,h) = d(h) $$ where $$ \frac{|d(h)|}{||(d,h)||} \leq \|d\| \frac{1}{2}\xrightarrow[(d,h) \to 0]{} 0$$