Let $H$ be a $\mathbb R$-Hilbert space, $x\in H$ with $\left\|x\right\|_H=1$, $\alpha\in\mathbb R$, $$\mathbb H_{x,\:\alpha}:=\left\{y\in H:\langle x,y\rangle_H\ge\alpha\right\}$$ and $U$ be an open subset (in the subspace topology) of $\mathbb H_{x,\:\alpha}$.
Now let $E$ be a $\mathbb R$-Banach space. $f:U\to E$ is called $C^1$-differentiable at $u\in U$ if $$\left.f\right|_\Omega=\left.\tilde f\right|_\Omega\tag1$$ for some $\tilde f\in C^1(O,E)$ for some $H$-open neighborhood $O$ of $u$ and $\Omega:=O\cap U$.
How can we show that $${\rm D}f(u):={\rm D}\tilde f(u)\tag2$$ is well-defined, i.e. independent of the choice of $\tilde f$?
I've read this claim in a lecture note, but it is not clear to me why it holds. It's trivial when $u\in\mathbb H_{x,\:\alpha}^\circ$ though.
Let $(O_1,\tilde f_1)$ and $(O_2,\tilde f_2)$ be two possible choices for $(O,\tilde f)$ and $\varepsilon>0$. Then there is a $\delta_i>0$ with $$\frac{\left\|\tilde f_i(v)-\tilde f_i(u)-{\rm D}\tilde f_i(u)(v-u)\right\|_E}{\left\|v-u\right\|_H}<\frac\varepsilon2\tag3$$ for all $v\in O_i$ with $0<\left\|v-u\right\|_H<\delta_i$. Let \begin{align}O&:=O_1\cap O_2,\\\Omega&:=\Omega_1\cap\Omega_2=O\cap U\end{align} and $\delta:=\min(\delta_1,\delta_2)$. Then it's clear that $$\frac{\left\|\left({\rm D}\tilde f_1(u)-{\rm D}\tilde f_2(u)\right)(v-u)\right\|_E}{\left\|v-u\right\|_H}<\varepsilon\tag4$$ for all $v\in\Omega$ with $0<\left\|v-u\right\|_H<\delta$. But why is this enough to conclude ${\rm D}\tilde f_1(u)={\rm D}\tilde f_2$?
It's clear such $v$ exist: Since $U$ is $\mathbb H_{x,\:\alpha}$-open, $$U=V\cap\mathbb H_{x,\:\alpha}\tag5$$ for some open subset $V$ of $H$. Since $O\cap V$ is $H$-open, there is a $\rho\in(0,\delta]$ with $$B_\rho(u)\subseteq O\cap V\tag6.$$ Now, if $$v:=\frac\rho2x+u,$$ then $$\langle x,v\rangle_H>\alpha$$ and hence we even got $$v\in\Omega^\circ=O\cap V\cap\mathbb H_{x,\:\alpha}^\circ\tag7.$$
Maybe everything is simpler than I thought: Let $O$, $\Omega$, $\tilde f$ and $V$ be as in the question. Since $O\cap V$ is $H$-open, there is a $\rho>0$ with $$B_\rho(u)\subseteq O.$$ Note that $$\Omega=O\cap V\cap\mathbb H_{x,\:\alpha}.$$ Let $h\in H$. Then, $${\rm D}\tilde f(u)h={\rm D}_h\tilde f(u)=\lim_{t\to0}\frac{\tilde f(u+th)-f(u)}t\tag8.$$ Assume $\left\|h\right\|_H=1$. Then, $$u+th\in B_\rho(u)\;\;\;\text{for all }|t|<\rho\tag9.$$ Note that $$\langle x,u+th\rangle_H\ge\alpha+t\langle x,h\rangle_H\;\;\;\text{for all }t\in\mathbb R\tag{10}.$$ If $\langle x,h\rangle\ge0$, then $$u+th\in B_\rho(u)\cap\mathbb H_{x,\:\alpha}\subseteq\Omega\;\;\;\text{for all }t\in[0,\rho)\tag{11}$$ and if $\langle x,h\rangle\le0$, then $$u+th\in B_\rho(u)\cap\mathbb H_{x,\:\alpha}\subseteq\Omega\;\;\;\text{for all }t\in(-\rho,0]\tag{12}.$$ So, if $\langle x,h\rangle\ge0$, then $${\rm D}\tilde f(u)h=\lim_{t\to0+}\frac{f(u+th)-f(u)}t\tag{13}$$ and if $\langle x,h\rangle\le0$, then $${\rm D}\tilde f(u)h=\lim_{t\to0-}\frac{f(u+th)-f(u)}t\tag{14}.$$
This should be enough to conclude or am I missing something?