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As part of determining the expression for the gradient in terms of an arbitrary inner product, I arrived at the following problem: Given:

  • $y = (y^1, \dots, y^n) \in \mathbb{R}^n$ is a selected point
  • $h = (h^1, \dots, h^n) \in \mathbb{R}^n$ is an arbitrary point.
  • $f$ is a real-valued function defined on $\mathbb{R}^n$
  • $[g_{ij}]$ is a positive definite, symmetric $n \times n$ matrix

I'm trying to show that

$$ \sum\limits_{j=1}^n \partial_{j}f(x_0)h^j = \sum\limits_{j,k=1}^n g_{jk}y^j h^k $$


$$ \partial_{j} f(x_0) = \sum\limits_{k=1}^n g_{jk}y^k $$

The only way I can reasonably see how to arrive at the conclusion is to reason as follows: Since the antecedent holds for arbitrary points in $\mathbb{R}^n$ it must hold for the particular $n$ points associated with the standard basis vectors $e_1, \dots, e_n$ So, for example, if we take the first point $h = (1, 0, \dots, 0)$ it follows that $\partial_{1}f(x_0) = \sum\limits_{k=1}^n g_{1k}y^k$ and so on and hence for the $j^{th}$ point we have the conclusion above.

So, my questions:

  • Is this line of reasoning correct?
  • Is there a better or more direct way to demonstrate the conclusion?


I'm updating this post to provide additional contextual information.

Given that $[g_{ij}]$ is a positive-definite and symmetric $n \times n$ matrix, it can be shown that the function

$$ (. | . )^g:\mathbb{R}^n \rightarrow \mathbb{R} $$

given by

$$ (x | y)^g = \sum\limits_{j,k=1}^n g_{jk}y^j x^k $$

is a scalar product on $\mathbb{R}^n$. Now, let $f$ be a function on $\mathbb{R}^n$ that is differentiable at $x_0$. By the Riesz represenation theorem (for finite-dimensional Hilbert spaces) since $df(x_0)$ is a continuous linear form there exists a unique vector $y$ such that $df(x_0)h = (y | h)^g \; \forall h\in \mathbb{R^n}$

This unique vector $y$ is defined to be the gradient of $f$ at $x_0$ with respect to the scalar product $(x | y)^g$ and is denoted by $y = \nabla^g f(x_0)$ I am working through the details of the proof that

$$ \nabla^g f(x_0) = (g^{1k}\partial_{k}f(x_0), \dots, g^{nk}\partial_{k}f(x_0)) $$

where the repeated upper/lower indices indication summation from $1 \dots n$ and $g^{ij}$ represents the $i-j$ entry of the inverse of the matrix $[g_{ij}]$

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Yes, your line of reasoning is correct. – Jesse Madnick May 31 '11 at 21:43
What you're saying seems correct, but I'm a bit puzzled because of your motivation. Don't you want to show that there exists a unique $y$ such that $df(x_0) h = g(y,h)$ for all $h$, so how can $y$ be selected? To see the existence and uniqueness of $y$ simply note that it must be $g$-orthogonal to the $(n-1)$-dimensional kernel of $df(x_0)$ (if the kernel is $n$-dimensional then $y$ must be zero by positive-definiteness of $g$), so choose such a vector and scale it appropriately. – t.b. May 31 '11 at 21:54
@Theo I have updated the post to provide additional information so that the context of the question will hopefully be more clear. – ItsNotObvious May 31 '11 at 22:48
I'm still puzzled. You can just take your candidate for $\nabla^g f(x_0)$ and check that $df(x_0)h = g(\nabla^g f(x_0),h)$ for all $h$ which is what you want and rather tautologically true. Or do you want to see how to find this candidate? Be that as it may, I'd say you've got everything you need already worked out :) – t.b. May 31 '11 at 23:19
Hi 3Sphere: Did you get where I'm heading? – t.b. Jun 1 '11 at 15:49

If you implant $${\rm grad}f\cdot h^{\top}=yG\cdot h^{\top}$$ for all $h$, then $$({\rm grad}f-yG)\cdot h^{\top}=0$$

so ${\rm grad }f=yG$ is exactly what you are claiming.

The qualities of $Q$ makes no difference.

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