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After seeing the integral definitions of div, grad, and curl, I'm left to wonder if we can define the regular derivative of a function $f: \Bbb R \to \Bbb R$ as the limit of an integral. For reference, here's the integral definition of grad $$\operatorname{grad} f := \lim_{V\to 0} \frac{\oint_S f(x)d\sigma}{V}$$ where $S$ is the boundary of the volume $V$.

In this same vein, is the following definition also correct? $$\frac{df}{dx}:=\lim_{b\to a} \frac{\int_a^bf(x)dx}{b-a}$$

where $f$ is continuous in $[a,b]\subseteq \Bbb R$. If so, how could it be proven?

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In your definition of the regular derivative, you should have an integral over the boundary of $[a,b]$, not over $[a,b]$ itself. The boundary of $[a,b]$, is just the set $\{a,b\}$ (with an orientation attached). We define the integral over this set as $f(b)-f(a)$ (minus sign because of that orientation business). Thus the integral definition should be $$\lim_{b\to a} \frac{f(b)-f(a)}{b-a}$$ But this is just the regular definition of the derivative.

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If $f$ is continuous, then your right-hand side converges to $f(a)$ by some MVT for integrals. The correct analog is $$ Df(a) = \lim_{b \to a} \frac{f(b)-f(a)}{b-a}, $$ where $f(b)-f(a)$ is the... zero-dimensional integral on $\{a,b\}$ of $f$.

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I'll add this, since I was just thinking about something similar.

If we have a function $g$ such that $\int_a^b g(t)dt=0$ and $\int_a^b g^2(t)dt=A>0$ then we have that

$$ f'(x)=\lim_{h \rightarrow0} \frac{1}{hA} \int_a^b f( x+h\cdot g(t))\cdot g(t)~dt. $$

To prove this, add and subtract $f(x) g(t)$ inside the integral, move things around a tad, multiply by $g(t)/g(t)$ on one part, and interchange the limit and integral on that part. So as long as we are able to do that, and $f$ and $g$ are nice enough, then this should hold.

$$\begin{align} \lim_{h \rightarrow0} \frac{1}{hA} &\int_a^b f( x+h\cdot g(t))\cdot g(t)-f(x)g(t)+f(x)g(t)~dt\\ &=\frac{1}{A} \int_a^b \lim_{h \rightarrow0} \frac{f( x+h\cdot g(t)) - f(x)}{h\cdot g(t)} \cdot g^2(t)~dt+ \lim_{h \rightarrow0}\frac{1}{hA} \int_a^b f(x)g(t)~dt\\ &=\frac{1}{A} \int_a^b f'(x)\cdot g^2(t)~dt+ \lim_{h \rightarrow0}\frac{1}{hA} \int_a^b f(x)g(t)~dt\\ &=f'(x)\cdot \frac{1}{A}\int_a^b g^2(t)~dt + f(x)\cdot \lim_{h \rightarrow0}\frac{1}{hA} \int_a^b g(t)~dt\\ &=f'(x)\cdot 1 + f(x)\cdot \lim_{h \rightarrow0}\frac{1}{hA} \cdot 0\\ &=f'(x) \end{align}$$

Basically we are using the function $g$ to look near $x$ and sort of averaging out the values of $f$ and then taking the limit as we look infinitesimally left and right. So it really sounds a lot like a derivative.

I imagine there are many ways to do this. It looks sort of like Cauchy's integral formula. I got the idea for this from the definition of curl as the limit of a circulation integral.


Here is another generalization. Let $(X,\Sigma,\mu)$ be a measure space, $g,h: X\rightarrow \mathbb R$ and $D\subset X$ such that $\int_D h~d\mu=0$ and $\int_D g \cdot h~d\mu=A>0$. Then

$$ f'(x)=\lim_{\epsilon \rightarrow0} \frac{1}{\epsilon A} \int_D f( x+\epsilon\cdot g(t))\cdot h(t)~\mu(dt). $$

The standard derivative definition can be gotten with $X=\{a,b\}$, $\mu( \{ i \} )=1$, $h(a)=-1$, $h(b)=1$, $g(a)=0$, $g(b)=1$. Then $\int_D g\cdot h~d\mu=1=A$ and $$\begin{align} \frac{1}{\epsilon A} \int_D f( x+\epsilon\cdot g(t))\cdot h(t)~\mu(dt) & = \frac{1}{\epsilon A} \Big( f( x+\epsilon\cdot g(a))\cdot h(a)\cdot \mu(\{a\})+f( x+\epsilon\cdot g(b))\cdot h(b)\cdot \mu(\{b\})\Big) \\ & = \frac{1}{\epsilon } \Big( -1 \cdot f(x+\epsilon\cdot 0) + 1 \cdot f(x+\epsilon)\Big) \\ & =\frac{f(x+\epsilon) - f(x)}{\epsilon} \end{align}$$ and clearly taking the limit as $\epsilon\rightarrow0$ gives $f'(x)$.

I'm sure there are certain requirements in order for this to be a general statement, but it shouldn't be too restrictive.

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No, it is not. By Lebesgue's differentiation theorem, for any $f\in L^1(\mathbb{R})$ and for almost every $x\in\mathbb{R}$ the following limit exists and it equals $f(x)$: $$ \lim_{r\to 0}\frac{1}{2r}\int_{|t-x|<r}f(t)\,dt.$$ If $f$ is a continuous function it is absolutely continuous over any compact interval, hence: $$ \lim_{b\to a}\frac{1}{b-a}\int_{a}^{b}f(t)\,dt = f(a), $$ i.e. any point is a Lebesgue point for a continuous function.


On the other hand, it is also true that for any entire function $f(z)$ we have:

$$ f'(z) = \frac{1}{2\pi i}\oint_{\|w-r\|=r}\frac{f(w)}{(z-w)^2}\,dw, $$ that is Cauchy's integral formula. Moreover, it is very common to define (pseudo)differential operators through particular integrals. See, for instance, the Wikipedia page about fractional calculus.

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