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As I began to teach myself in differential geometry, I finally used to use the Nabla-Operator.

I know and understand its usage as in

$$ \nabla f := \left( \begin{matrix} \frac{∂f}{∂x_1} & \frac{∂f}{∂x_2} & \cdots & \frac{∂f}{∂x_n} \end{matrix} \right)^\intercal $$

but in many books I read a pure definition of $\nabla$: $$ \nabla := \left( \begin{matrix} \frac{∂}{∂x_1} & \frac{∂}{∂x_2} & \cdots & \frac{∂}{∂x_n} \end{matrix} \right) ^ \intercal $$

which seems to be just a visualisation of the content, because it's mathematically false $-$ an equation needs to have two evaluatable terms on both sides, but an operator is not a value.

For example, the derivation operator can conformly be defined as $$ \frac{∂}{∂x_i}: ℝ → ℝ, \quad f ↦ \frac{∂f}{∂x_i} := \lim_{x_i→0}{\frac{f(x_1,\cdots,x_i+h,\cdots,x_n)-f(_1,\cdots,x_i,\cdots,x_n)}{h}} $$

But the Nabla-Operator is applied in multiple ways; therefore, one cannot define it as a function.

Do I suppose rightly that there does not exists an explicit definition, or does there exist some kind of ‘trick’?

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  • $\begingroup$ What kind of function is the function $f$ that you want to apply the nabla operator to? $\endgroup$ – BIS HD Nov 15 '13 at 14:03
  • $\begingroup$ I don't really want to apply the nabla operator to anything, I just want to define it explicitly (if possible)… but in most cases, $ f: ℝ^3 → ℝ^3 $. $\endgroup$ – Luke Nov 15 '13 at 14:05
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    $\begingroup$ The differentiation operator maps the some space of functions into some other space of functions. It does not make any sense to write: $\frac{\partial}{\partial x_i}:\mathbb{R}\rightarrow\mathbb{R}$. $\endgroup$ – Sergio Parreiras Nov 15 '13 at 14:27
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    $\begingroup$ We can define $D: C^k(\mathbb{R})\rightarrow C^{k-1}(\mathbb{R})$ by $D(f)=f^\prime$ where $C^k$ is the set of functions that have a continuous k-th derivative. $\endgroup$ – Sergio Parreiras Nov 15 '13 at 15:23
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    $\begingroup$ Why can't you define $\nabla: C^k(\mathbb{R}^n)\rightarrow \left( C^k(\mathbb{R}^n)\right)^n$ by $\nabla \left(f\right)=(\frac{\partial}{\partial_1}f,\ldots,\frac{\partial}{\partial_n}f)$ ? $\endgroup$ – Sergio Parreiras Nov 15 '13 at 21:12
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The keyword is Operator Calculus or alternatively Operational Calculus. Here is an introductory (PDF) document . Other references are easily found on the internet, such as Fractional Calculus (Wikipedia), What is operator calculus? (MSE), How to make sense of this calculus notation, Advanced College Level (MSE).

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I find this an interesting question, because it appeared to me also when I studied multivariable calculus. First of all, "an equation needs to have two evaluatable terms on both sides, but an operator is not a value" is meaningless (and also false, no matter what meaning you give to "evaluatable terms"). What does this even mean? Also $\nabla:=(\partial_1,\ldots,\partial_n)^T$ is not a mere equation, it is a definition. Your only objection can be to the meaning of the right hand sie.

"But the Nabla-Operator is applied in multiple ways; therefore, one cannot define it as a function. Do I suppose rightly that there does not exists an explicit definition, or does there exist some kind of ‘trick’?" Doesn't make any sense either, in particular "therefore, one cannot define it as a function" is a non sequitur. Counterexample: let $\Omega\subset\mathbf{R}^n$ be open, and $X$ be the set of all functions $\Omega\rightarrow\mathbf{R}$ such that $\partial_i f$ exists for all $i$. Also let $Y$ be the set of all maps $\Omega\rightarrow\mathbf{R}^n$. Then we define $\nabla:X\rightarrow Y$ by $\nabla f:=(\partial_1 f,\ldots,\partial_n f)^T$. It is a perfectly ok definition. You see, in the definition it is UNDERSTOOD (IMPLICIT) how the function $\nabla$ is to be defined (one does not need "operator calculus" or whatever), writing $\nabla:=(\partial_1,\ldots,\partial_n)^T$ already suggests this definition. The symbol $(\partial_1,\ldots,\partial_n)^T$ should not be understood as an element of $\mathbf{R}^n$ (because it isn't one), but in the sense I made precise.

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