Motivations
First Motivation: I got a “new mean” by indirectly attempting to establish the relationship between the uniform probability measure and normalized Hausdorff measure. This is the third open problem in Section 10, pg. 37 of this paper, “The Maximum Entropy of a Metric Space”:
The uniform measure (pg. 32-37), when defined, is a canonical probability measure on a given metric space. But so too is the Hausdorff measure. More exactly, if the Hausdorff dimension d of X is finite, then we have the Hausdorff measure \mathcal{H}^{d} on X, which if 0<\mathcal{H}^{d}(X)<\infty can be normalized to a probability measure. What is the relationship between the Hausdorff probability measure and the uniform measure? It is probably not simple: for example on \{1,1/2,1/3,\cdots,0\}\subseteq\mathbb{R}, the uniform measure is well-defined (it is \delta_0), but the Hausdorff probability measure is not.
(Since we are using computers, I decided to take the Lebesgue measure and simple functions defined “almost everywhere” on \mathbb{R} rather than the Hausdorff measure and sophisticated functions defined on “fractal subsets” of \mathbb{R}.)
I used the open problem to compute “the extended mean w.r.t. the Lebesgue measure”: the extended Lebesgue mean is defined after the motivations.
Regardless, I want to solve a different question, addressing the problem with computing the Cauchy principal value of asymmetrical functions.
Second Motivation: I assume the Cauchy principal value and “extended Lebesgue mean” of a function (after the motivation) are equivalent, when the function is symmetric around its singularity. However, when the function is asymmetric around its singularity is where things get interesting.
The problem with the Cauchy principal value is it oversimplifies the mean around the singularity of a function. If we integrate over all sequences of bounded functions with subdomains of positive and finite Lebesgue measure converging to the domain of asymmetrical f, the mean can be any value.
For instance, suppose f:[-2,0]\setminus\{-1\}\to\mathbb{R} and f(x)=1/(x^3+1). The Cauchy principal value is the following (i.e., b_1(r)=1/r, b_2(r)=1/r, and x_s=-1 is the singularity of f(x)):
Clear["Global`*"]
f[x_] := f[x] = 1/(x^3 + 1)
bound1[r_] := bound1[r] = -1/r (*b1[r]*)
bound2[r_] := bound2[r] = 1/r (*b2[r]*)
s = -1 (*singularity in f*)
CauchyPrincpleValue =
Limit[(1/((s -
bound1[r] - (-2)) + (0 - (s + bound2[r])))) (Integrate[
f[x], {x, -2, s - bound1[r]},
Assumptions -> {r \[Element] Reals && r > 2}] +
Integrate[f[x], {x, s + bound2[r], 0},
Assumptions -> {r \[Element] Reals && r > 2}]), r -> Infinity]
(*Output: 1/36 (-Sqrt[3] \[Pi] + 6 Sqrt[3] ArcTan[5/Sqrt[3]] + Log[343])*)
However, when we change b_1(r) such that b_1(r)=-1/(2r) and b_1(r)\neq b_2(r), then we have the following (note the Log terms in the output of both codes are non-equivalent):
Clear["Global`*"]
f[x_] := f[x] = 1/(x^3 + 1)
bound1[r_] := bound1[r] = -1/r (*b1[r]*)
bound2[r_] := bound2[r] = 1/r (*b2[r]*)
s = -1 (*x-value of singularity in f*)
CauchyPrincpleValue =
Limit[(1/((s -
bound1[r] - (-2)) + (0 - (s + bound2[r])))) (Integrate[
f[x], {x, -2, s + bound1[r]},
Assumptions -> {r \[Element] Reals && r > 2}] +
Integrate[f[x], {x, s + bound2[r], 0},
Assumptions -> {r \[Element] Reals && r > 2}]), r -> Infinity]
(*Output: 1/36 (-Sqrt[3] \[Pi] + 6 Sqrt[3] ArcTan[5/Sqrt[3]] + Log[343/64])*)
Since f has no symmetry on [-2,0]\setminus\{-1\} and the Cauchy principal value does not consider the different rates of b_1(r) and b_2(r), we can conclude the Cauchy principal value is unreliable for this and other unbounded asymmetrical functions.
Definition and Attempt to Answer the Second Motivation
Consider the following solution that chooses between the multiple rates of changes between subdomains of f converging to its domain rather than intervals of equal length.
Definition 1 (The Extended Lebesgue Mean).
Suppose,
f:A\to\mathbb{R} is a function such that A\subseteq\mathbb{R} is a subset
\lambda(\cdot) is the Lebesgue Measure defined on the Borel \sigma-algebra,
the integral is defined, w.r.t. the Lebesgue measure (i.e., the Lebesgue integral);
|\cdot| is the absolute value,
\mathbf{R}:=(\mathbf{R}_1,\mathbf{R}_2),
(x,f(x))\in\mathrm{graph}(f),
\mathrm{sign}(\cdot) is the sign function,
\mathbb{G}(\mathbf{R},f)=\mathrm{sign}(f(x)-\mathbf{R}_2)\sqrt{|x-\mathbf{R}_1|^2+|f(x)-\mathbf{R}_2|^2},
\mathrm{I} is the appropriate index set of a family of sets
\mathbb{B}(A) is the set of all families of bounded sets such that \{B_v:v\in\mathrm{I}\}\in\mathbb{B}(A) whenever
- B_v\subseteq A for all v\in\mathrm{I},
- \lambda(B_v)\in(0,+\infty) for all v\in\mathrm{I},
- \bigcup_{v\in\mathrm{I}}B_v=A,
- \exists(\mathbf{R}\in\mathbb{R})\forall(\varepsilon_1>0)\exists(N_1\in\mathrm{I})\forall(v\in\mathrm{I})\\ \left(v\ge N_1\Rightarrow\left|\dfrac{1}{\lambda(B_v)}{\displaystyle\int_{B_v}}\mathbb{G}(\mathbf{R},f)\, d\lambda\right|<\varepsilon_1\right).
\dfrac{1}{\lambda(B_v)}{\displaystyle\int_{B_v}}\mathbb{G}(\mathbf{R},f)\, d\lambda\to 0 \text{ as } v\to\inftyThe extended Lebesgue mean of f: A\to\mathbb{R} is M_f(A) whenever the following is true:
\begin{aligned} & \forall(\{A_r:r\in\mathrm{I}\}\in\mathbb{B}(A))\exists!(M_{f}(A)\in\mathbb{R})\forall(\epsilon>0)\exists(N\in\mathrm{I})\forall(r\in\mathrm{I}) \\ &\left(r\ge N\Rightarrow\left|\frac{1}{\lambda(A_r)}\int_{A_r}f\, d\lambda-M_{f}(A)\right|<\epsilon\right). \end{aligned}
Question
Can “the extended Lebesgue mean” of an asymmetric f with a singularity exist (e.g., f:[-2,0]\setminus\{-1\}\to\mathbb{R} such that f(x)=1/(x^3+1)) which also gives a different average from the Cauchy principal value of f?
Attempt to Solve Question
I assume,
only when \mathbf{R}=(-1,0), but I am not sure how to confirm the statement.
I decided to try something more simple. Using the integral in the second motivation, we take the following:
Suppose x_s+b_1(r) and x_s+b_2(r) are rates of change around the singularity x_s of f(x)=1/(x^3+1) (i.e., x_s=-1). We check for any b_1(r)\to 0^{-} and any b_2(r)\to 0^{+} such that when:
\begin{align} \mathbb{G}(\mathbf{R},f)= & \,\mathrm{sign}\left(f(x)-\mathbf{R}_2\right)\sqrt{|x-\mathbf{R}_1|^2+\left|f(x)-\mathbf{R}_2\right|^2}\\ =&\, \mathrm{sign}\left(\frac{1}{x^3+1}-0\right)\sqrt{|x-(-1)|^2+\left|\frac{1}{x^3+1}-0\right|^2}\\ = & \, \mathrm{sign}\left(\frac{1}{x^3+1}\right)\sqrt{|x+1|^2+\left|\frac{1}{x^3+1}\right|^2} \end{align}and
(Criterion 1): \,{\dfrac{1}{(x_s-b_1(r)-(-2))+(0-(x_s+b_1(r)))}\left({\displaystyle\int_{-2}^{x_s-b_1(r)}}\mathbb{G}(\mathbf{R},f)+{\displaystyle\int_{x_s+b_2(r)}^{0}}\mathbb{G}(\mathbf{R},f)\right)=0}then any b_1(r)\to 0^{-} and b_2(r)\to 0^{+} that satisfies criterion 1. has to give the same value for:
{\frac{1}{(x_s-b_1(r)-(-2))+(0-(x_s+b_1(r)))}\left(\int_{-2}^{x_s-b_1(r)}\frac{1}{x^3+1}+\int_{x_s+b_1(r)}^{0}\frac{1}{x^3+1}\right)}to give hope the extended Lebesgue mean M_f([-2,0]\setminus\{-1\}) exists and answers the question.
There might be a chance one could use Mathematica to confirm the blockquote. (Perhaps this is impossible, since we need to find all pairs of functions b_1(r)\to 0^{-} and b_2(r)\to 0^{+} that satisfy criterion 1.)
Either way, here are two examples I assume might give the same mean (edit: they don’t give the same mean but their integrals have the same value they give the same mean):
- b_1(r)=1/(tr) and b_2=1/r such that t=5.536933 whenever r=100,000,000 so we have criterion 1. has the smallest absolute difference from zero. (The code below does not work; however, it can be fixed. The error is below
Criterion1[t]andNewMean. Note, I found these results using Desmos.)
Clear["Global`*"]
f[x_] := f[x] = 1/(x^3 + 1)
R1 = -1
R2 = 0
R = {R1, R2}
bound1[t_, r_] := bound1[t, r] = 1/(t r) (*b1[r]*)
bound2[r_] := bound2[r] = 1/r (*b2[r]*)
s = -1 (*x-value of singularity in f*)
G[x_] := G[x] = Sign[f[x] - R2] EuclideanDistance[{R1, R2}, {x, f[x]}]
(*The function G(R,f)*)
Criterion1[t_] :=
Criterion1[t] =
Limit[(1/((s -
bound1[t, r] - (-2)) + (0 - (s + bound2[r])))) (Integrate[
G[x], {x, -2, s + bound1[t, r]},
Assumptions -> {r \[Element] Reals && r > 2}] +
Integrate[G[x], {x, s + bound2[r], 0},
Assumptions -> {r \[Element] Reals && r > 2}]), r -> Infinity]
(*The Integral of G(R,f) in Criterion 1*)
tsolve = NSolve[Criterion1[t] == 0, t]
(*Using Desmos, I approximated tsolve to 5.53693..., but I do not know if
this is correct. In Mathematica, it returns the following error.*)
(*Error: Integrate::idiv: Integral of Sqrt[Abs[1+x]^2+1/Abs[1+x^3]^2]/
Sign[1+x^3] does not converge on {-2,-1-1/(r t)}.*)
NewMean =
Limit[(1/((s -
bound1[tsolve,
r] - (-2)) + (0 - (s + bound2[r])))) (Integrate[
f[x], {x, -2, s + bound1[tsolve, r]},
Assumptions -> {r \[Element] Reals && r > 2}] +
Integrate[f[x], {x, s + bound2[r], 0},
Assumptions -> {r \[Element] Reals && r > 2}]), r -> Infinity]
(*Using Desmos, I get the answer is .165907; however, in Matheamtica I get
an internal error invovling iCopy(expression)*)
- b_1(r)=1/r and \require{enclose}\enclose{horizontalstrike}{b_2=1/(r^v)} b_2=1/(rv) such that \require{cancel}\xcancel{v=\frac{10,078,094}{11,111,111}} v=.18060582 whenever r=100,000,000 so that we have criterion 1. has the smallest absolute difference from zero. (The code below also does not work, but could be fixed. The error is below
Criterion1[v]andNewMean. I found these results using Desmos.)
Clear["Global`*"]
f[x_] := f[x] = 1/(x^3 + 1)
R1 = -1
R2 = 0
R = {R1, R2}
bound1[r_] := bound1[r] = 1/ r (*b1[r]*)
bound2[r_, v_] := bound2[r, v] = 1/(rv) (*b2[r]*)
s = -1 (*x-value of singularity*)
G[x_] := G[x] = Sign[f[x] - R2] EuclideanDistance[{R1, R2}, {x, f[x]}]
(*The function G(R,f)*)
Criterion1[v_] :=
Criterion1[v] =
Limit[(1/((s -
bound1[r] - (-2)) + (0 - (s + bound2[r, v])))) (Integrate[
G[x], {x, -2, s - bound1[r]},
Assumptions -> {r \[Element] Reals && r > 2}] +
Integrate[G[x], {x, s + bound2[r, v], 0},
Assumptions -> {r \[Element] Reals && r > 2}]), r -> Infinity]
(*The Integral of G(R,f) in Criterion 1*)
vsolve = NSolve[Criterion1[v] == 0, {v, 1}]
(*Using Desmos, I approximated vsolve to .18060582, when r=100,000,000.
There is a possibility that vsolve is Infinity. Hence, we have to make
adjustments to bound2[r,v]. In Mathematica, it returns the following
error.*)
Integrate::idiv: Integral of Sqrt[Abs[1+x]^2+1/Abs[1+x^3]^2]/
Sign[1+x^3] does not converge on {-1+rv,0}.
NewMean =
Limit[(1/((s -
bound1[r] - (-2)) + (0 - (s +
bound2[r, vsolve])))) (Integrate[
f[x], {x, -2, s - bound1[r]},
Assumptions -> {r \[Element] Reals && r > 2}] +
Integrate[f[x], {x, s + bound2[r, vsolve], 0},
Assumptions -> {r \[Element] Reals && r > 2}]), r -> Infinity]
(*Using Desmos, I get the answer is .082954; however, in Mathematica I
get an internal error invovling iCopy(expression)*)
Using Desmos, I get both codes return the integrals of NewMean in both codes is \require{enclose}\enclose{horizontalstrike}{.165907}, but since the total lengths of their intervals around the singularities are different, their means are different.
Sub-Question: Is there any cases besides a symmetrical function, where the extended Lebesgue mean of an unbounded elementary function exists?
Using Desmos, I get both codes generate NewMean to be .082954. This is different from the Cauchy Princpal Value of f which is .736387.
Clear["Global`*"]
CauchyPrincipalValue=N[Integrate[1/(x^3 + 1), {x, -2, 0}, PrincipalValue
-> True]]
(*Output: .736387*)
Sub-Question: Is .082954 the extended Lebesgue mean of f:[-2,0]\setminus\{-1\}\to\mathbb{R} such that f(x)=1/(x^3+1)?