Something strange about uniqueness of the derivative in higher dimensions

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  • #1
quasar987
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Main Question or Discussion Point

Recall that for a function [itex]f:A\subset \mathbb{R}^n\rightarrow \mathbb{R}^m[/itex], the derivative of f at x is defined as the linear map L:R^n-->R^m such that ||f(x+h)-f(x)-L(h)||=o(||h||)
if such a linear map exists.

We can show that for certain geometries of the set A, when the derivative exists, it is unique. This is the case for instance if A is a closed disk or more simply, if A is any open set. However, for certain sets, the derivative may exists and not be unique. For instance, if A is a singleton, then any linear map does the trick.

I arrive at a strange conclusion if I assume that a function [itex]f:A\subset \mathbb{R}^n\rightarrow \mathbb{R}^m[/itex] is differentiable at x with L_1, L_2 two distinct derivatives of f at x and follow the proof that the matrix representation of the derivative is the Jacobian matrix.

By definition, we have that [itex]||f(x+h)-f(x)-L_k(h)||=o(||h||) [/itex] (k=1,2), which implies by the sandwich theorem that [itex]|f_j(x+h)-f_j(x)-(L_k)_j(h)_|=o(||h||)[/itex], for each component j=1,...,m. In particular,

[tex]\lim_{t\rightarrow 0}\left|\frac{f_j(x_1,...,x_i+t,...,x_n)-f_j(x_1,...,x_n)-(L_k)_j(0,...,t,...,0)}{t} \right|=0 [/tex]

from which it follows by definition of the partial derivatives that

[tex]\frac{\partial f_j}{\partial x_i}(x)=(L_k)_j(e_i) [/tex]

But this is absurd since L_1 and L_2 are assumed distincts so there is at least a couple (i,j) for which [itex](L_1)_j(e_i)\neq (L_2)_j(e_i) [/itex], leading to the contradiction

[tex]\frac{\partial f_j}{\partial x_i}(x)\neq \frac{\partial f_j}{\partial x_i}(x) [/tex]

Does anyone sees where I'm mistaken in my reasoning??
 

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  • #2
HallsofIvy
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In using that definition for partial derivatives (in particular that limit), you are assuming that domain is an open set- one of the conditions under which the derivative is unique.
 
  • #3
quasar987
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Doesn't it suffice to define the partial derivative at x [itex](\partial f/\partial x_i)(x) [/itex] of a function [itex]f:A\subset \mathbb{R}^n\rightarrow\mathbb{R}^m[/itex] as

[tex]
\lim_{t\rightarrow 0}\frac{f(x_1,...,x_i+t,...,x_n)-f(x_1,...,x_n)}{t}
[/tex]

that there exists in A\{x} a sequence [tex]\{(x_1,...,y_k,...,x_n)\}_{k\geq 1}[/tex] with y_k-->x_i ??
 
  • #4
quasar987
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Nevermind the definition of partial derivative; we have for each i=1,...,n, j=1,...,m and k=1,2,

[tex]
\lim_{t\rightarrow 0}\left|\frac{f_j(x_1,...,x_i+t,...,x_n)-f_j(x_1,...,x_n)-(L_k)_j(0,...,t,...,0)}{t} \right|=0
[/tex]

Or equivalently,

[tex]
\lim_{t\rightarrow 0}\frac{f_j(x_1,...,x_i+t,...,x_n)-f_j(x_1,...,x_n)}{t}=(L_k)_j(0,...,1,...,0)
[/tex]

But since L_1 and L_2 are different, there exists at least a couple (i,j) such that [tex](L_1)_j(0,...,1,...,0)\neq (L_2)_j(0,...,1,...,0)[/tex], leading to the contradiction mentionned, without mentioning the definition of partial derivatives.
 
  • #5
HallsofIvy
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Doesn't it suffice to define the partial derivative at x [itex](\partial f/\partial x_i)(x) [/itex] of a function [itex]f:A\subset \mathbb{R}^n\rightarrow\mathbb{R}^m[/itex] as

[tex]
\lim_{t\rightarrow 0}\frac{f(x_1,...,x_i+t,...,x_n)-f(x_1,...,x_n)}{t}
[/tex]

that there exists in A\{x} a sequence [tex]\{(x_1,...,y_k,...,x_n)\}_{k\geq 1}[/tex] with y_k-->x_i ??
No, it doesn't. It is possible for a function to have all partial derivatives (at a point) and yet not be differentiable at that point.

For example, the function f(x,y)= 1 if [itex]xy\ne 0[/itex], 0 if xy= 0, has both partial derivatives equal to 0 at (0,0) but is not differentiable there.

It is possible to prove that if a function has all partial derivatives in an open set, and those partial derivatives are continuous on that set, then the function is differentiable in the set.
 
  • #6
quasar987
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Mmmh, maybe I'm misunderstanding the issue here because I don't understand the point you wanted to make with your last post.

Or maybe you misunderstood my post no.3? Post no.4 was an attempt at disambiguation of post no.3.

Thanks Halls for the discussion.
 
  • #7
matt grime
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I can't see what you're getting at, quasar.

You have shown that if a function has all its partial derivatives, and that they're continuous at x, and defined on an open neighbourhood of x, then there is a unique derivative there. That is just what Halls stated.
 
  • #8
quasar987
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Where have I shown that?

What I believe I did in post no.1 is to assume that f was differentiable at a point x of a (non-open) set A with L_1, L_2 two distinct derivatives of f at x. Then I wrote the definition of the statement "L_i is a derivative of f at x" and looked at the special case h=(0,...,t,...,0) from which a contradiction with the hypothese that the derivative is not unique is apparent.
 
  • #9
mathwonk
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since a derivative is supposed to be linear map that approximates the given function, it makes little sense to discuss derivatives unless the domain of the original function is a space that can be approximated by a linear space, i.e. a manifold. so singleton spaces are approximated by the linear space {0}, in which a linear map IS unique.

i.e. derivatives are unique whenever they make sense.
 

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