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It says we can define gradient as the unique vector $\nabla f$ such that $Df(x)(v)=\langle \nabla f(x),v \rangle$

I don't understand how uniqueness is coming. I can prove uniqueness if it was given $dim(Hom(E,W))$ is finite where $f:E\subset V \to W$ and differentiable on open subset $E$ of $V$.

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3 Answers 3

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Use the fact that if $w_1, w_2$ are elements of an inner product space such that $\langle w_1, v\rangle = \langle w_2, v\rangle$ for all $v$, then $w_1 = w_2$.

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  • $\begingroup$ yes that'll follow from non degeneracy but here $w_1=\nabla f_1,w_2=\nabla f_2$ right ? $\endgroup$
    – dragoboy
    Sep 4, 2014 at 2:30
  • $\begingroup$ There is no $f_1$ and $f_2$. The function $f$ is fixed. We are supposing that there are two vectors which satisfy the property which defines $\nabla f(x)$. $\endgroup$ Sep 4, 2014 at 3:33
  • $\begingroup$ Can you please write full proof ? I'm not at all getting this thing, I need to understand it. Like, first of all how $\nabla f(x)$ is coming and then why unique $\endgroup$
    – dragoboy
    Sep 4, 2014 at 8:10
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    $\begingroup$ $Df(x)$ is a linear map on a finite dimensional inner product space, so it is of the form $Df(x)(v) = \langle w, v\rangle$ for some $w$; this is an application of the Reisz representation theorem. The vector $w$ is unique as I alluded to in my answer; we call the vector $w$ the gradient of $f$ at $x$, that is, we set $\nabla f(x) = w$. $\endgroup$ Sep 4, 2014 at 22:02
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The problem appears to be assuming that $Df(x)$ is a linear map from a vector space V with an inner product <,> to $\mathbb{R}$. Correct, yes? If so, then think about the connection between linear maps $V \rightarrow \mathbb{R}$ and the inner product. It helps to look at some more concrete cases for these problems. For example, where V is just Cartesian space $\mathbb{R}^n$. If $L:V \rightarrow \mathbb{R}$ is linear, what would a formula for $L(x_1,x_2, x_3, ...)$ look like? Well, $$L(x_1,x_2,x_3) = 2x_1+x_2-3x_3$$.

Can you express the right-hand side as an inner product of $(x_1,x_2,x_3)$ and some other vector $c$? What would $c$ be if $L(x_1,x_2,x_3)$ is to be $<c,x>$?

Now think about the original question. $Df(x)$ simply denotes some linear map like $L$. Must there always be formula for $L$ like the one in the example?

Start with the above considerations and continue until done ;-)

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Found this question while thinking about the same problem as well. Actually this directly follows from Riesz representation theorem, which states that every linear functional can be uniquely represented by an inner product with a vector.

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