Let $X\subset E$ where $E$ is a Banach space, and $[\alpha,\beta]\subset\Bbb R$. Then for $x\in X$ we define $u_x$ as the constant map defined by $[\alpha,\beta]\to E,\,t\mapsto x$. Then for the function $\psi$ defined by $x\mapsto u_x$ we clearly have $\psi\in C^\infty(X,C([\alpha,\beta],X))$ where $\partial\psi(x)h=u_h$.
I dont understand how is differentiated $\psi$ to obtain $\partial\psi$, that is, $\psi(x)=u_x$ but, how to differentiate it? Can someone show me, step by step, how we can obtain (formally) this result? Thank you in advance.
By the definition of Frechet derivative, we shall find a bounded linear operator $A : X \to C([\alpha,\beta],X)$ such that $$\lim_{h \to 0} \frac{ \| \psi(x + h) - \psi(x) - Ah \|_{C([\alpha,\beta],X)} }{ \|h\|_{X} } = 0.$$
Now, by the definition of $\psi$ this can be rewritten as $$\lim_{h \to 0} \frac{ \| u_{x + h} - u_{x} - Ah \|_{C([\alpha,\beta],X)} }{ \|h\|_{X} } = 0.$$
Here, for $t \in [\alpha,\beta]$ we have $$(u_{x + h} - u_{x} - Ah)(t) = (x+h) - x - Ah = (1-A)h$$ so, since $\|\cdot\|_{C([\alpha,\beta],X)} = \|\cdot\|_u,$ the uniform norm defined by $\|f\|_u = \sup_x \|f(x)\|_X,$ we have $$\|u_{x + h} - u_{x} - Ah\| = \sup_{t \in [\alpha,\beta]} \|(1-A)h\|_X = \|(1-A)h\|_X.$$
Thus we shall have $$\lim_{h \to 0} \frac{ \|(1-A)h\|_X }{ \|h\|_{X} } = 0.$$
This implies that $\|1-A\|_{L(X, C([\alpha,\beta], X))} = \sup_{h \in X} \frac{\|(1-A)h\|_X}{\|h\|_X}$ must be zero, i.e. $A = 1.$
We set $\partial\psi(x) = A.$ This gives $\left(\partial\psi(x)h\right)(t) = Ah = h = u_h(t),$ i.e. $\partial\psi(x)h = u_h.$