Definition of differentiability on closed sets

Question

Let $U\subset \mathbb{R}^{n}$ be an open set. A function $f: U \to \mathbb{R}^{m}$ is said to be differentiable at $a \in U$ if there exists some linear map $Df(a):\mathbb{R}^{n}\to \mathbb{R}^{m}$ such that: $$\lim_{h\to 0}\frac{||f(a+h)-f(a)-Df(a)(h)||}{||h||} = 0 $$ This is standard. However, during a course, my professor mentioned that we can define differentiability when $U$ is not assumed to be open as follows. If $U$ is not necessarily open, then $f$ is said to be differentiable at $a \in U$ if there exists some open set $V\subset \mathbb{R}^{n}$ and a function $\tilde{f}: V \to \mathbb{R}^{m}$ such that $a \in V$, $\tilde{f}$ is differentiable at $a$ and $\tilde{f}\bigg{|}_{V\cap U} \equiv f$.

I understand this definition. But isn't it possible that there are two different open sets $V_{1}$ and $V_{2}$ and functions $\tilde{f}_{1},\tilde{f}_{2}$ satisfying the above conditions? And if so, shouldn't $\tilde{f}_{1}\bigg{|}_{V_{1}\cap U} = \tilde{f}\bigg{|}_{V_{2}\cap U} \equiv f$? It does not seem to be the case since these functions seem not to have the same domain.

Answer 1

Yes, it is possible that there are two different open sets $V_1,V_2$ and functions $\tilde{f}_1,\tilde{f}_2$ satisfying these conditions. In fact, it's usually the case that if such extensions exist at all, there's plenty of them. Take for example $U=\mathbb{R}\times\{0\}\subset\mathbb{R}^2$, $f\colon U\rightarrow\mathbb{R}^2,\,(x,0)\mapsto(x,0)$. Then you can pick $V_1=V_2=\mathbb{R}^2$ (and any other open set containing $U$ works as well) and $\tilde{f}_1=\operatorname{id}_{\mathbb{R}^2}$, $\tilde{f}_2\colon\mathbb{R}^2\rightarrow\mathbb{R}^2,\,(x,y)\mapsto(x,0)$.

But there is a small error in your definition. You demand $\tilde{f}\vert_{V\cap U}=f$, but the function on the LHS is defined on $V\cap U$ and the function on the RHS is defined on $U$. So, instead, one should require $\tilde{f}\vert_{V\cap U}=f\vert_{V\cap U}$ and all will be well.

If we now have two such extensions, we will indeed not necessarily have $\tilde{f}\vert_{V_1\cap U}=\tilde{f}\vert_{V_2\cap U}$, because these functions can have different domains. However, we will, by definition, always have $\tilde{f}_1\vert_{V_1\cap V_2\cap U}=f\vert_{V_1\cap V_2\cap U}=\tilde{f}_2\vert_{V_1\cap V_2\cap U}$.

Answer 2

I argue there is one piece of information that might be missing for other StackExchangers seeing this post. Note that the definition of differentiability implies

$$\lim_{(y-x)\rightarrow 0} \frac{\|\tilde f_i(x) - \tilde f_i(y) - \mathrm{D}\tilde f_i|_y(x-y)\|}{\|x-y\|} =0$$

$$\forall u \in \mathcal{S}^{n-1}_1, \quad \lim_{t\rightarrow 0} \frac{\|\tilde f_i(y+tu) - \tilde f_i(y) - \mathrm{D}\tilde f_i|_y tu\|}{\|tu\|} =0$$

$$\forall u \in \mathcal{S}^{n-1}_1, \quad \lim_{t\rightarrow 0} \left\|\frac{\tilde f_i(y+tu) - \tilde f_i(y)}{t} - \mathrm{D}\tilde f_i|_y u\right\| =0$$

$$\forall u \in \mathcal{S}_1^{n-1}, \quad \lim_{t\rightarrow 0} \frac{\tilde f_i(y+tu) - \tilde f_i(y)}{t} = \mathrm{D}\tilde f_i|_yu.$$

In the case when $U$, on which $f:U\rightarrow \mathbb{R}^m$, is closed then, as mentioned by Thorgott,

$$\tilde{f}_1|_{V_1\cap V_2\cap U} = \tilde{f}_2|_{V_1\cap V_2\cap U} = {f}|_{V_1\cap V_2\cap U}.$$ We are then only really concerned with directions $u$ where

$$\forall \epsilon \in[0,\infty), \exists t \in [0,\infty) : ut \in \mathcal{S}^n_\epsilon \cap U \subseteq V_1\cap V_2 \cap U.$$

Call this set $\mathcal{U}$. Turns out for these directions the "derivative" is unique:

$$\forall u \in \mathcal{U}, \quad \mathrm{D}\tilde f_1|_yu = \lim_{t\rightarrow 0} \frac{\tilde f_1(y+tu) - \tilde f_1(y)}{t} = \lim_{t\rightarrow 0} \frac{\tilde f_2(y+tu) - \tilde f_2(y)}{t} = \mathrm{D}\tilde f_2|_yu.$$