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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 $V$.

Michael Albanese
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dragoboy
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3 Answers3

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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 $w_1 = w_2$.

Michael Albanese
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  • yes that'll follow from non degeneracy but here $w_1=\nabla f_1,w_2=\nabla f_2$ right ? – dragoboy Sep 04 '14 at 02:30
  • 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 $\nabla f(x)$. – Michael Albanese Sep 04 '14 at 03:33
  • 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 – dragoboy Sep 04 '14 at 08:10
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    $Df(x)$ 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 $\nabla f(x) = w$. – Michael Albanese Sep 04 '14 at 22:02
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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.

Rui Liu
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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 ;-)

kmiker
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