Definition talk:Gradient Operator/Real Cartesian Space

Isn't the gradient supposed to be a vector-valued function rather than a column matrix? Also, as for now, the gradient has only been defined at points $\mathbf x$. --Lord_Farin 10:14, 2 April 2012 (EDT)
 * ooh, I got confused about our discussion in my sandbox page. Let me try to fix it. --GFauxPas 10:17, 2 April 2012 (EDT)
 * Okay so looking for closely at Larson he says $\nabla$ is an operator in the same sense $\frac {\mathrm d}{\mathrm dx}$ is an operator. He says that $\nabla$ operates on $f(\mathbf x)$ and produces the vector $\nabla f(\mathbf x)$. I think he means $\nabla: \R^\R \to \R^n, f \mapsto \nabla f$. In any event what I put down is what Larson has as a definition, he has this stuff as a footnote --GFauxPas 10:26, 2 April 2012 (EDT)


 * I would say we have $\nabla: [\R^n\to\R] \to [\R^n\to\R^n]$ where $[X\to Y]$ is intended to be $Y^X$, the set of all mappings (of course, the domain has to be suitably restricted). So, it transforms functions into functions. But $\nabla$ surely does not operate on $f\left({\mathbf x}\right)$ because $f(x) = g(x)$ does not imply $\nabla f(x) = \nabla g(x)$, evidently. Thus, $\nabla$ needs to be seen as taking functions as input, giving again functions as output. --Lord_Farin 10:33, 2 April 2012 (EDT)


 * I don't understand your explanation as to how we know $\nabla$ doesn't operate on $f(\mathbf x)$ but I think you're right. I'm having a hard time getting clarity because too many sources use $f$ to mean $f(\mathbf x)$ and vice versa which in general is annoying but here is particularly annoying. I think what's happening here is indeed that $f \mapsto \nabla f$. But we were able to define "derivative" without coming on to $(f: x \mapsto y) \mapsto (D_xf: x \mapsto \frac {\mathrm dy}{\mathrm dx})$. I'm guessing many people are more comfortable with "functions on numbers/vectors" than "functions on functions". But then again, It is standard fare to learn $f + g$, $f \circ g$ etc. in high school... --GFauxPas 10:54, 2 April 2012 (EDT)


 * In fact, one could define first the gradient at a point, not insisting that all partial derivatives of $f$ exist everywhere. If this definition is applied at every point, we naturally obtain a mapping; if you like, you can view $\nabla$ as a function taking as input $f$ and $\mathbf x$, but this is of course clearly equivalent to $\nabla$ assigning a function $\nabla f$ as value to $f$. I tried to make clear that the value $\nabla f (\mathbf x)$ does not depend solely on $f$ through the value $f (\mathbf x)$. In this sense, $\nabla$ needs 'more information about $f$' to be able to determine $\nabla f$ (namely, the behaviour of its derivatives); in this sense, $\nabla$ isn't merely a function $\R\to\R^n$ taking $f(\mathbf x)$ as an argument. --Lord_Farin 13:07, 2 April 2012 (EDT)

It should be $S^{\R^n}$ instead of $S^S$, or rather $\left[{S \to \R^n}\right]$ (but I am not sure this latter notation was already introduced on PW; it's just very convenient sometimes). --Lord_Farin 14:03, 2 April 2012 (EDT)
 * My meta-ing wasn't meta enough! Feel free to put up the notation for us? --GFauxPas 14:04, 2 April 2012 (EDT)
 * Wait, $S^{\R^n}$ or ${\R^n}^S$? --GFauxPas 14:07, 2 April 2012 (EDT)
 * The latter; my bad. 't is done, see Definition:Set of All Mappings. --Lord_Farin 14:14, 2 April 2012 (EDT)
 * Still more to whine: $\nabla$ isn't defined on $[S\to\R]$ but on the subset thereof of mappings $f$ for which $\nabla f$ exists. --Lord_Farin 14:22, 2 April 2012 (EDT)
 * You lost me. Doesn't the way we defined $S$ avoid that problem? --GFauxPas 15:04, 2 April 2012 (EDT)
 * Well, it solves the problem for $S$. However, $[S \to \R]$ also contains functions which are plainly not continuous. The problem lies in the fact that $S$ depends implicitly on $f$ (so it is sort of $S_f$ in some sense), but there may be functions $g$ in $[S_f \to \R]$ for which $S_g \subsetneq S_f$ or even $S_g = \varnothing$. So, $[S\to\R]$ should be restricted to an appropriate subset of functions whose $\nabla$ is defined on all of $S$. Hopefully, this sheds some light. --Lord_Farin 15:11, 2 April 2012 (EDT)
 * New edit covers exactly my point; thought may go out to a more elegant description of $\mathbf F$, but for the moment it suffices. --Lord_Farin 15:23, 2 April 2012 (EDT)