1
$\begingroup$

Let's consider the function $f:\mathbb R^n\rightarrow\mathbb R$. According to Wikipedia, the gradient of $f$ is defined as the unique vector field whose dot product with a unit vector $\mathbf v$ is the directional derivative of $f$ in the direction of $\mathbf v$: $$D_\mathbf v f = \nabla f\cdot\mathbf v.$$ I've been wondering how do we know that such a vector field exists and is unique.

Any help is appreciated, thanks.

$\endgroup$
1
  • $\begingroup$ The title asks or uniqueness, which is trivial - just let $\mathbf v$ run over a basis $\endgroup$ – Hagen von Eitzen Aug 15 '20 at 19:38
1
$\begingroup$

Given any linear map $F\colon\Bbb R^n\longrightarrow\Bbb R$, there is one and only one vector $w\in\Bbb R^n$ such that$$(\forall v\in\Bbb R^n):F(v)=v.w.$$You just take$$w=\bigl(F(e_1),F(e_2),\ldots,F(e_n)\bigr),$$where $\{e_1,e_2,\ldots,e_n\}$ is the standard basis.

If you apply this theorem to $D_vf$, you get the existence and the unicity of the gradient.

$\endgroup$
0
0
$\begingroup$

If you use definition with the canonical basis you have that $\nabla f \cdot e_j=d_j f$ where $e_j$ is the j-th canonical vector and $d_j$ indicates the $j$-th partial derivate. So in canonical basis the gradient needs to be the vector that has the $j$-th partial derivate in the $j$-th position. From this the uniqueness.

$\endgroup$
0
$\begingroup$

For another approach, simply note that, on fixing $x\in \mathbb R^n$, one has $\nabla f(x)\cdot v=f'(x)v$, and now uniqueness follows from the definition of derivative, as soon as the latter is known to exist.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.