Fundamental Theorem

Let $Df(x)=f(x+1)-f(x)$ denote the discrete derivative of a continuous function f on the real line. In this post, I assume that all functions are continuous of have compact support. They are zero outside some large interval. No smoothness is required of course. Here is the simplest version of the fundamental theorem of calculus in a discrete setting:

Theorem: $\int f'(x) dx = 0$ .

Proof. Choose a constant $R$ so that a $r$ neighborhood of the support of $f$ is included in the open interval $(-R,R)$ . Using substitution we have $\int_{-R}^{R} f(x+r)-f(x-r) \; dx = \int_{-R+r}^{R+r} f(x) \; dx-\int_{-R-r}^{R-r} f(x) \; dx =0$

This version of the fundamental theorem is also true for periodic functions. We think then as the function defined on the circle.

As in the classical case, we can use this to switch the differentiation in a product. Note first the Leibniz differentiation rule $D(f g) = f(x+1) g(x+1) - f(x) g(x) = Df(x) g(x+1) + f(x) Dg(x)$

which leads to the integration by parts formula where we use the notation $f_-(x)=f(x-1)$ :

Corollary: $\int f g'(x) \; dx = - \int f'_- g \; dx$ .

This is also a consequence of the fact that the backwards derivative $f \to f'(x-1)=f(x)-f(x-1)$ is the adjoint of the forward derivative $f \to f'(x) = f(x+1)-f(x)$ . Now think of $\int f g dx$ as an inner product. Also this corollary is the same for periodic functions.

weierstrass function and derivative — Weierstrass function and its derivative

Define the anti-derivative of a function of compact support as $Sf(n) = \sum_{k < n} f(k)[/latex] where [latex]k[/latex] runs over the integers. Here are the two versions of the fundamental theorem of calculus: <b>Theorem:</b> [latex]D S(f)(x) = S D(f)(x) = f(x)$

They contain the essence of calculus, are as deep and useful as the classical versions and are so simple that they could have been found thousands of years ago, because all what is needed is basic arithmetic, no Riemann or even Lebesgue integration theory is needed. The theorem tells that taking differences and summing up are inverses operations of each other. Difference the sum or sum the difference and you end up with the same function. For functions without compact support we can define $Sf(n) = \sum_{0 \leq k < n} f(k)[/latex] and get [latex]D S(f)(x) = f(x), S D(f)(x) = f(x)-f(0) [/latex]. We have met the functions [latex][x]^n$ already in the Taylor series post. Since $D[x]^n = n [x]^{n-1}$ we have $S[x]^n=[x]^{n+1}/(n+1)$ . This allows to integrate any function of compact support by developing the function into a Taylor series. Note that for all continuous functions of compact support, the Taylor series $\sum_{n=0}^{\infty} D^nf(a) [x-a]^n/n!$ converges for all $a$ . For functions of compact support, it makes sense to chose a below outside the support of $f$ . Then

Corollary: $Sf(x) = \sum_{n=1}^{\infty} D^nf(a) [x-a]^{n+1}/(n+1)!$ .

This looks exactly as in the classical case. But this formula is true for all continuous functions of compact support. Not even continuity is needed. While this result is just playing with summation and differences, it is in essence known since antiquity. The only thing which is new now is that the formalism is done in such a way that it looks like the result in classical calculus. This comes handy because we do not want to remember anything new when dealing with discrete calculus structures. The lack of good notation is usually the biggest enemy of discrete structures. I’m sure that if you have studied numerical methods, especially in partial differential equation you can sing a song about bad notation.

Define the boundary of an interval $I=[a,b]$ as the union of intervals $I' =[a,a+1] \cup [b,b+1]$ . We assume that $|b-a|>1$ . Here is an other version of the fundamental theorem:

Theorem: $\int_I f'(x) dx = \int_{I'} f(x) dx$ .

Also this statement only suffers from the defect that it is too obvious to be appreciated, even so it contains the essence of the classical version. Here is the proof: $\int_a^b f'(x) dx=\int_a^b f(x+1)-f(x) dx=\int_{a+1}^{b+1} f(x) dx - \int_a^b f(x) dx=\int_I f(x) dx$ .

Lets say that a continuous function is harmonic if f”(x)=0. As in the classical case all harmonic functions of compact support must be zero:

Corollary: All harmonic functions are zero

Proof. On any sequence $\dots a-3,a-2,a-1,a,a+1,a+2, \dots$ the function is linear and so zero. Alternatively, just use the Taylor formula and integrate the function 0 twice so that $S(S(0)) =0$ .

Author: oliverknill

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