Uniqueness of high-dimensional derivative [Zorich's book] I was reading the definition of the high-dimensional derivative from Zorich's book and I'd like to ask a question about the uniqueness of high-dimensional derivative.

Definition 1. A function $f:E\to \mathbb{R}^n$ defined on a set $E\subset \mathbb{R}^m$is differentiable at the point $x\in E$,
which is a limit of $E$, if  $$f(x+h)-f(x)=L(x)h+\alpha(x;h), \quad \quad \quad(1)$$ where
$L(x):\mathbb{R}^{m}\to \mathbb{R}^n$ is a function that is linear in
$h$ and $\alpha(x;h)=o(h)$ as $h\to 0, x+h\in E$.
The linear function $L(x):\mathbb{R}^m\to \mathbb{R}^n$ in $(1)$ is
called the  differential of the function $f:E\to
 \mathbb{R}^n$ at the point $x\in E$.

Claim. If $E\subset \mathbb{R}^m$ and $f:E\to\mathbb{R}^n$ is differentiable at the point $x\in E$, which is a limit point of $E$, then the differential of $f$ at the point $x\in E$ is unique.
Proof. Let's assume there are $A,B:\mathbb{R}^m\to \mathbb{R}^n$ linear mappings such that
$$f(x+h)-f(x)=Ah+\alpha(x;h) \quad \text{and} \quad f(x+h)-f(x)=Bh+\beta(x;h),$$ where $\alpha(x;h)=o(h)$ and $\beta(x;h)=o(h)$ as $h\to 0, x+h\in E$.
Remark. Here the base is $h\to 0, x+h\in E$. The elements of this base are $B_{\delta}:=\{h\in \mathbb{R}^m: 0\leq \lVert h\rVert<\delta, x+h\in E\}$ for $\delta>0$. Indeed, it satisfies to the definition of base.
Let $C:=A-B$, then $C:\mathbb{R}^m\to \mathbb{R}^n$ is a linear mapping also. Then it is not difficult to show that $$\lim \limits_{\substack{h\to 0 \\ x+h\in E}} \frac{\lVert Ch\rVert}{\lVert h\rVert}=0.$$
Let $y$ be a nonzero vector in $\mathbb{R}^m$. If we can show that $\lim \limits_{\substack{\mathbb{R}\ni t\to 0 \\ x+ty\in E}} \frac{\lVert C(ty)\rVert}{\lVert ty\rVert}=0,$ then we are done. Indeed, in that case it follows that $\lVert Cy\rVert=0$ for all $y\neq 0$, which implies that $Cy=0$ for all $y$ and hence $Ay=By$. Therefore, linear mappings $A$ and $B$ are equal.
Question. How to prove that if $\lim \limits_{\substack{h\to 0 \\ x+h\in E}} \frac{\lVert Ch\rVert}{\lVert h\rVert}=0$, then $\lim \limits_{\substack{\mathbb{R}\ni t\to 0 \\ x+ty\in E}} \frac{\lVert C(ty)\rVert}{\lVert ty\rVert}=0$ ??
It is certainly true if $x$ is an interior point of $E$ but it does not look correct to me because we claim that $x$ is the limit point of $E$. Let me try to explain why do I think so.
Let $\varepsilon>0$ be given. Take $\delta_0=\frac{\delta}{\lVert y\rVert}>0$, where $\delta>0$ comes from the first limit. Then for any $t\in \mathbb{R}$ such that $0<|t|<\delta_0,\ x+ty\in E$ we'll get that $\frac{\lVert C(ty)\rVert}{\lVert ty\rVert}<\varepsilon$. It seems that we are done but how do we know that such $t$ exists?
I mean how do we know that $\{t\in \mathbb{R}: 0<|t|<\delta_0, x+ty\in E\}\neq \varnothing$?
Thank you so much for your help!
 A: Such a $t$ does not necessarily exist. You can take a hyperplane $E$ passing through the origin $ x = 0$ and $y$ any vector which spans a line which is not included in the hyperplane $E$. In this case one has
$$ x + ty \not \in E \quad  \forall t \in \mathbb R_0. $$
In this case we do not have unicity of the differential. Indeed consider $f = k \in \mathbb R$ constant on $E$.  Then $$ f(x + h) - f(x) = 0.$$
Hence it follows that any linear map $L_x : \mathbb R^m \to \mathbb R : h \mapsto L_x(h)$ whose restriction to the hyperplane $E$ is equal to $0$ satisfies the definition of differentiability.


Your example is showing that if we assume that the function $f:E\to \mathbb R^n$ is differentiable at the point $x\in E$, where $x$ is a limit point of E, then differential of f at $x$ may not be unique, right?

Yes. However if the limit point $x$ happens to be an interior point then we do have unicity.

To avoid this ambiguity one needs to assume that x is an interior point of E or E is an open set, right?

Not really. Asking that $x$ be an interior point garanties the unicity. But this is not needed in order to have unicity. Indeed it is possible ins some cases for $x$ to be a limit point, $x$ not an interior point and still have  unicity. That is, $x$ being an interior point is a sufficient condition but not needed.
Example
Let $E = [0,1]^2 \subset \mathbb R^2$ and $f = k \in \mathbb R$ be constant on $E$. Define
$$ L_x = 0 : \mathbb R^2 \rightarrow \mathbb R$$
Then for all $x  + h = h \in E$
$$f(x+h)-f(x) - L(x)h = 0 = o(h),$$
so $f$ is differentiable at $x = 0$. To prove the unicity suppose there are other linear maps $A,B$ satisfying the definition then let $C = A-B$ and we have
$$ \lim_{h\to 0, x + h \in E} \frac{\Vert Ch \Vert}{\Vert h \Vert} = 0.$$
Let $y_1 = (1,0), y_2 = (0,1) \in E$. Then for all $0 < t < 1$ we have
$ x + ty_i \in E$ hence we have
$$ \lim_{t \to 0 , t > 0 }  \frac{\Vert C(ty_i) \Vert}{\Vert ty_i \Vert} = 0.$$
In particular we deduce that $$ A(y_i) = B(y_i).$$
Since the two linear maps $A$ and $B$ coincide on a basis of the vector space $\mathbb R^2$ it follows that $$ A = B.$$

If we consider one-dimensional case, i.e. f:E→R where E⊂R and f is differentiable at x∈E, then in that setting we always have unicity of differential even x is a limit point of E, right? I think the subtle difference here is that in R we have only one direction

If we adapt our proof to the case where $E$ is of dimension one then yes that's the general idea. However for the case where $E \subset \mathbb R$ there is a much simpler argument to prove the unicity. Indeed there are linear maps $L_1,L_2$ and $\alpha_i(x;h) = o(h)$ such that
$$ f(x + h) - f(x) = L_i(x)h + \alpha_i(x;h)$$
then
$$ \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = L_i(x) $$
so $L_1 = L_2$ by unicity of the limit.
