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)

Differentiability
How is this connected to differentiability at a point? Is the gradient operator used if there is no tangent (hyper)plane? --Dfeuer (talk) 04:26, 31 January 2013 (UTC)


 * The gradient is defined at every point where the derivative exists. Not necessarily conversely because "approaching diagonally" may yield different limits. This area needs ample attention from someone with time and a good source on multivariate real analysis. --Lord_Farin (talk) 08:03, 31 January 2013 (UTC)