# How to prove the uniqueness of the gradient?

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.

• The title asks or uniqueness, which is trivial - just let $\mathbf v$ run over a basis Aug 15, 2020 at 19:38

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.
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.
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.