Suppose $f(x)$ where $x \in D \subset R^n$ is a real-valued function. Why is the gradient of $f(x)$, i.e. $\nabla f(x)$, or subgradient of $f(x)$, i.e. $g \in \partial f(x)$, in the dual space?
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2$\begingroup$ The gradient at a point is usually a vector which lies in the tangent space. The total derivative $df$ at that point, on the other hand, is a covector in the cotangent space. $\endgroup$– Alex ProvostMay 3, 2018 at 19:52
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$\begingroup$ Could you explain your notion through a mathematical fashion? $\endgroup$– user494522May 3, 2018 at 21:44
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$\begingroup$ I think to have a better answer one should define primal space and dual space, then discusses why they are orthogonal. $\endgroup$– user494522May 3, 2018 at 21:49
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1 Answer
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The gradient is a map $C^{1}\left(D,\,\mathbb{R}\right)\times D\to L\left(D,\,\mathbb{R}\right)$, where $L$ denotes the space of linear maps. (Strictly speaking we would only need differentiable and not $C^{1}$, but meh...) So if you fix a function $f$ and an evaluation point $x$ (where you take the gradient), you get an element of $L\left(D,\,\mathbb{R}\right)$ which is the dual space.
The trick is actually to forget all that you learned about derivatives in highschool. Derivatives are not numbers, partial derivatives are. The gradient gives you for each spot in the domain of a function a linear map describing the behaviour of the function at that point.
answered May 3, 2018 at 19:53
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$\begingroup$ What is $C^1$ and ... in your answer? In addition, we know that the gradient is a vector but why it lives in the dual space and is perpendicular to the primal space? $\endgroup$– user494522May 3, 2018 at 21:46
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$\begingroup$ $C^1$ means once continuously differentiable and strictly the continuity of the derivative is not required $\endgroup$ May 3, 2018 at 22:40
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$\begingroup$ Also as I was attempting to explain, the notion of the gradient as a "vector" in the context of functional analysis is somewhat missing the point: Here we think of derivatives as mappings to the dual space, its simply a definition. (The vector concept is useful in numerics, for example). $\endgroup$ May 3, 2018 at 22:46
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$\begingroup$ Also, if you want an easy explanation of the orientation of the vector associated with the gradient: We write the "primal" vectors as columns. In order to have a scalar-valued liner map on the "primal"-space, we thus need a row-vector. As the gradient is supposed to be such a map, it needs to be a row vector. - Alex Provost's comment is actually "mathematically" better, but also requires more background. $\endgroup$ May 3, 2018 at 22:53
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$\begingroup$ I think to be complete we should state why $L(D, \mathbb{R})$ is not in general in the original space? $\endgroup$– mathtickAug 9, 2021 at 9:49