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As a simple example, suppose we have a function $f: \mathbb{R}^3 \to \mathbb{R}$ defined on the set (and taking $+\infty$ everywhere else),

$$\{x \in \mathbb{R}^3| x_1 \in [-1, 1], x_2 \in [-1, 1], x_3 = 0\}$$

The set has no interior but a relative interior given by $(-1,1) \times (-1,1) \times \{0\}$.

Similarly, consider sets such as $\{x \in \mathbb{R}^3| \langle e, x\rangle = 1, x_i \geq 0\}$, where $e$ is the one-vector. Once again, it has no interior, but has a relative interior relative to the hyperplane $\langle e, x \rangle = 1$ given by $\{x \in \mathbb{R}^3| \langle e, x\rangle = 1, x_i > 0\}$,

Example functions could include:

$f(x) = \langle x, x \rangle$ for the first set

$f(x) = -\langle e, \ln(x) \rangle$ for the latter set

Are such function differentiable on such sets (i.e. the gradient exists)? If not, why? Can't seem to find any resource on this.

Edited per comment: Also, is it problematic if I were to pretend that the function was defined on the whole space, take the gradient there, and restrict it to the relative interior? For example, consider $f(x) = \langle x, x \rangle$ defined on the set $[-1, 1]^2 \times \{0\}$. What is wrong if I were to take the gradient as usual, $\nabla f(x) = 2 x$ and define it on the relative interior of the same set $(-1, 1)^2 \times \{0\}$?

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  • $\begingroup$ It's not clear to me what you are asking. Are you asking for a function that is defined on all of $\mathbb R^3$ (as the notation $f : \mathbb R^3 \to \mathbb R$ indicates) but that is differentiable only on a set such as $[-1, 1]^2 \times \{0\}$, or are you looking for a definition of differentiability that works for functions defined only on $[-1, 1]^2 \times \{0\}$? $\endgroup$
    – LSpice
    Jul 27, 2020 at 21:17
  • $\begingroup$ If the latter, it may be interesting to consider the notion of manifolds with boundary, of which both of your spaces are natural examples. Here is the first Google result for me for "smooth functions on manifolds with boundary": math.ucr.edu/~res/math260s10/manwithbdy.pdf . $\endgroup$
    – LSpice
    Jul 27, 2020 at 21:20
  • $\begingroup$ In your second example $f(x) = -\langle e, \ln(x)\rangle$, do you mean to define $\ln(x_1, x_2, x_3) = (\ln(x_1), \ln(x_2), \ln(x_3))$? If so, then how do you define the function on the boundary? $\endgroup$
    – LSpice
    Jul 27, 2020 at 21:23
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    $\begingroup$ @LSpice I guess a even lower example is whether a function, e.g., $f(x) = x_1^2 + x_2^2$ defined only for straight lines, e.g., it is only finite on the line $x_1 + x_2 = 1$ and $+infinity$ everywhere else. Is the function differentiable on that line in $\mathbb{R}^2$ (which also has no interior). Is the function differentiable. in the usual sense? And if there is a compatible definition in the case it is not. $\endgroup$ Jul 27, 2020 at 21:36
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    $\begingroup$ @Cauchy'sCarrot: I did not mean this to be offensive. To me, a question "let $f$ be infinite except at a single point, is it differentiable" makes little sense if "differentiable" is understood in the usual sense, because the definition of differentiability simply makes no sense in this case. You can speak about directional derivatives, though. $\endgroup$ Jul 28, 2020 at 0:24

2 Answers 2

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I think what you are looking for is the standard definition of once-differentiable manifold with boundary.

In order to define derivative, you need a normed vector space. You need a vector space because differentiation is ${\Bbb R}$-linear. You want to preserve the linearity because in a sense, differentiation is linearization. You need the norm to take the limit.

The concept is generalized to manifolds by looking at the infinitesimal neighborhood of a point, the tangent space, which is a vector space.

As LSpice commented, your examples are manifolds with boundary. The boundary is itself of a manifold one dimension lower, so you can define derivative there.

You can also consider a boundary point as part of the whole space. There your "tangent space" is only half of a vector space. You can generalize linearity here also, if you like.

Finally, manifolds are define by charts and you want to make sure that your differentiation operator is defined consistently across the charts. This means that the transition maps should be differentiable.

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You can parametrize such sets and then consider the differentiability with respect to the parameters. The differentiability property will be invariant with respect to diffeomorphisms: if two parametrizations are related by a diffeomorphism (that is, by a differentiable bijection whose inverse is also differentiable), then a function differentiable with respect to one of the two parametrizations will be differentiable with respect to the other parametrization. In general, a parametrization can be any bijection. Some parametrizations may be more useful/natural than others -- e.g., parametrizations that are homeomorphisms with respect to the natural topologies would usually be better than parametrizations that are not homeomorphisms.

E.g., you can parametrize the set $S:=\{x\in\mathbb R^3\colon x_1+x_2+x_3=1,x_i>0\ \forall i\}$ by the parametrization $$S_1\ni(s,t)\mapsto\phi(s,t):=(s,t,1-s-t)\in S$$ or, e.g., the parametrization $$S_2\ni(s,t)\mapsto\psi(s,t):=(1-s-t,s,t)\in S,$$ where $S_1:=S_2:=\{(s,t)\in\mathbb R^2\colon s>0,t>0,s+t<1\}$. These two parametrizations are equivalent, in the sense that they are related by a diffeomorphism -- here, specifically, by the diffeomorphism $$S_2\ni(s,t)\mapsto g(s,t):=(1-s-t,s)\in S_1$$ in the sense that $\psi=\phi\circ g$ and hence $\phi=\psi\circ g^{-1}$.

$\big($In the above example, the domains $S_1$ and $S_2$ of the two different parametrizations $\phi$ and $\psi$ of the same set $S$ were the same. In general, though, the domains of different parametrizations of the same set may be different. Even in the above example, another parametrization of $S$ is $$S_3\ni(s,t)\mapsto\rho(s,t):=(s,t-s,1-t)\in S,$$ where $S_3:=\{(s,t)\in\mathbb R^2\colon0<s<t<1\}\ne S_1$. The parametrization $\rho$ is then equivalent to the parametrizations $\phi$ and $\psi$.$\big)$

A function $f\colon S\to\mathbb R$ may then be called differentiable if the function $f\circ\phi\colon S_1\to\mathbb R$ is differentiable or, equivalently, if the function $f\circ\theta$ is differentiable, where $\theta$ is any parametrization of $S$ equivalent to $\phi$. Then, by the chain rule, we also have $$(f\circ\psi)'=(f\circ\phi\circ g)'=(f\circ\phi)'\circ g';$$ here, at each point of $S_2$, $g'$ is a linear operator from $\mathbb R^2$ to $\mathbb R^2$, and $(f\circ\phi)'$ is a linear operator from $\mathbb R^2$ to $\mathbb R$ (that is, a linear functional).

For instance, the function $S\ni x\mapsto f(x):=x_1^2+x_2x_3$ will be differentiable, because the function $S_1\ni(s,t)\mapsto (f\circ\phi)((s,t))=s^2+t(1-s-t)$ is differentiable or, equivalently, because the function $S_2\ni(s,t)\mapsto (f\circ\psi)((s,t))=(1-s-t)^2+st[=(f\circ\phi\circ g)((s,t))]$ is differentiable.

For further reading, see e.g. differentiation on manifolds.

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  • $\begingroup$ I'm not sure this fully answers the question, because your example is of a manifold without boundary, whereas the question explicitly wants to consider differentiation on manifolds with boundary. $\endgroup$
    – LSpice
    Jul 27, 2020 at 22:21
  • $\begingroup$ Is it problematic if I were to pretend that the function was defined on the entire set instead? For example, consider $f(x) = \langle x, x \rangle$ defined on the set $[-1, 1]^2 \times \{0\}$. What is wrong if I were to take the gradient as usual, $\nabla f(x) = 2 x$ and define it on the interior of the same set $(-1, 1)^2 \times \{0\}$? $\endgroup$ Jul 27, 2020 at 22:29
  • $\begingroup$ With this choice, the functions $f : x \mapsto x_1^2 + x_2^2 + x_3^2$ and $g : x \mapsto x_1^2 + x_2^2$, which agree on your given domain, would be found to have different gradients there, so that the gradient could no longer be said to be a property of the restriction of the function to $[-1, 1]^2 \times \{0\}$. $\endgroup$
    – LSpice
    Jul 27, 2020 at 22:39
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    $\begingroup$ @LSpice : Concerning your first comment, you are of course right. However, I thought the main idea should be first explained on the significantly easier case of a manifold without boundary. $\endgroup$ Jul 27, 2020 at 23:54
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    $\begingroup$ @Cauchy'sCarrot : As noted in the above comment by user LSpice, the gradient will not in general be a property of the restriction of the function. However, the two functions in the example given in that comment have in fact the same gradient, $(2x_1,2x_2,0)$, on the set $T:=(-1,1)^2\times\{0\}$. However, the functions $x\mapsto x_1+x_2+x_3$ and $x\mapsto x_1+x_2$ will have the same restriction to the set $T$, but their gradients, $(1,1,1)$ and $(1,1,0)$, will be different (everywhere and hence) on $T$. $\endgroup$ Jul 28, 2020 at 0:09

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