How to prove the uniqueness of the gradient?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.