D**u
发帖数: 204 | 1 On a 2-dim plane, F is a (real number valued) function on each polygon region
P. We also know that if P is the union of 2 disjoint polygon regions P1 and P2
, then
F(P) = F(P1) + F(P2).
Question: if for every rectangle D (no need to be parallel to x-y axis) we
have F(D) = 0, does that imply that F(P) = 0 for every polygon region P? |
s*******r
发帖数: 63 | 2 如果F处处可导的话，证明应该是简单的，把F看成一个积分即可。
如果有旋转和平移不变性的话，证明也是简单，只要证明所有对于说有直角三角形，F=
0，即可。 |
s***e
发帖数: 267 | 3 不需要这些假定。首先任何直角三角形都是0。不然两个直角三角形可以拼成一个矩形
，矛盾。
然后任何三角形可以分成两个直角三角形，所以也是0。任何polygon可以分成三角形。
F=
【在 s*******r 的大作中提到】
: 如果F处处可导的话，证明应该是简单的，把F看成一个积分即可。
: 如果有旋转和平移不变性的话，证明也是简单，只要证明所有对于说有直角三角形，F=
: 0，即可。
|
j******n
发帖数: 271 | 4 这里F可导怎么定义?
F=
【在 s*******r 的大作中提到】
: 如果F处处可导的话，证明应该是简单的，把F看成一个积分即可。
: 如果有旋转和平移不变性的话，证明也是简单，只要证明所有对于说有直角三角形，F=
: 0，即可。
|
g**********1
发帖数: 1113 | 5 为什么a+b=0 可以推出a=b=0?
【在 s***e 的大作中提到】
: 不需要这些假定。首先任何直角三角形都是0。不然两个直角三角形可以拼成一个矩形
: ，矛盾。
: 然后任何三角形可以分成两个直角三角形，所以也是0。任何polygon可以分成三角形。
:
: F=
|
j******n
发帖数: 271 | 6 There is a gap. F(P1) + F(P2) = 0 does not imply F(P1) = F(P2) = 0 for two
triangles P1 and P2 since F can be negative.
【在 s***e 的大作中提到】
: 不需要这些假定。首先任何直角三角形都是0。不然两个直角三角形可以拼成一个矩形
: ，矛盾。
: 然后任何三角形可以分成两个直角三角形，所以也是0。任何polygon可以分成三角形。
:
: F=
|
D**u
发帖数: 204 | 7 Just to make it clear, the problem does not assume "旋转和平移不变性".
F=
【在 s*******r 的大作中提到】
: 如果F处处可导的话，证明应该是简单的，把F看成一个积分即可。
: 如果有旋转和平移不变性的话，证明也是简单，只要证明所有对于说有直角三角形，F=
: 0，即可。
|
p*****k
发帖数: 318 | 8 DuGu, since both the area and the function are additive,
cannot we just use the Riemann sum definition of the area:
F(P)=F(sum dx dy)=sum F(dx dy)=0
then take limit? am i missing anything subtle? |
s*******r
发帖数: 63 | 9 由于已经有了additivity, “可导”类似于pdf的定义，大概的意思就是说，F可以写成
积分形式，F=\int_\poly f(x,y)dxdy，当然这里要求f(x,y) bounded。
同样的，一般情况下，f(x,y)可以是任意的，这样F就不能简单得取极限了。
前面我只是说，如果有这两个条件中的任意一个，就能够证明。否则，我不知道怎么证
，而且我也找不到反例。
【在 j******n 的大作中提到】
: 这里F可导怎么定义?
:
: F=
|
j******n
发帖数: 271 | 10 I guess we need some continuity such as:
lim_{n->\inf} F(P_n) = F(Union_{n from 0 to \inf} P_n), for P_n subset of
P_{n+1}.
region
and P2
we
【在 D**u 的大作中提到】
: On a 2-dim plane, F is a (real number valued) function on each polygon region
: P. We also know that if P is the union of 2 disjoint polygon regions P1 and P2
: , then
: F(P) = F(P1) + F(P2).
: Question: if for every rectangle D (no need to be parallel to x-y axis) we
: have F(D) = 0, does that imply that F(P) = 0 for every polygon region P?
|
|
|
j******n
发帖数: 271 | 11 pdf definition requires a stronger additivity: Countable additivity.
【在 s*******r 的大作中提到】
: 由于已经有了additivity, “可导”类似于pdf的定义，大概的意思就是说，F可以写成
: 积分形式，F=\int_\poly f(x,y)dxdy，当然这里要求f(x,y) bounded。
: 同样的，一般情况下，f(x,y)可以是任意的，这样F就不能简单得取极限了。
: 前面我只是说，如果有这两个条件中的任意一个，就能够证明。否则，我不知道怎么证
: ，而且我也找不到反例。
|
g**********1
发帖数: 1113 | 12 F只是一个function.
【在 p*****k 的大作中提到】
: DuGu, since both the area and the function are additive,
: cannot we just use the Riemann sum definition of the area:
: F(P)=F(sum dx dy)=sum F(dx dy)=0
: then take limit? am i missing anything subtle?
|
D**u
发帖数: 204 | 13 When taking limit, (if you want to prove F(P)=0) you still need to prove
F(P) = limit sum F(dx dy),
though you have
Area(P) = limit sum Area(dx dy).
【在 p*****k 的大作中提到】
: DuGu, since both the area and the function are additive,
: cannot we just use the Riemann sum definition of the area:
: F(P)=F(sum dx dy)=sum F(dx dy)=0
: then take limit? am i missing anything subtle?
|
p*****k
发帖数: 318 | 14
DuGu, i see - sorry that was a naive argument.
here are some thoughts based on what ppl have discussed so far:
since every polygon can be triangulated, and every triangle can be
decomposed into two right triangles, it suffices to prove this
"measure function" vanishes for all right-angled triangles.
now take the middle points of each side, along with the right-angled
vertex. this forms a rectangle; the rest are two similar triangles
of half of the original size. by repeating this, the origina
【在 D**u 的大作中提到】
: When taking limit, (if you want to prove F(P)=0) you still need to prove
: F(P) = limit sum F(dx dy),
: though you have
: Area(P) = limit sum Area(dx dy).
|
s*******r
发帖数: 63 | 15 I'm not sure whether "countable additivity" is sufficient, because F(d\sigma
) may not be 0.
【在 p*****k 的大作中提到】
:
: DuGu, i see - sorry that was a naive argument.
: here are some thoughts based on what ppl have discussed so far:
: since every polygon can be triangulated, and every triangle can be
: decomposed into two right triangles, it suffices to prove this
: "measure function" vanishes for all right-angled triangles.
: now take the middle points of each side, along with the right-angled
: vertex. this forms a rectangle; the rest are two similar triangles
: of half of the original size. by repeating this, the origina
|
D**u
发帖数: 204 | 16 The question does not require F to be countable additive.
In the traditional measure theory, every measurable set has a non-negative
measure, so if a measurable set S_1 is a subset of S_2, then measure(S_2) >=
measure(S_1). Then for a P in the 2-d plane, we can use a countable union
of rectangles to "approximate" the measure of P.
The same approximation fails in this question because F does not guarantee a
non-negative "measure" on any polygon.
【在 p*****k 的大作中提到】
:
: DuGu, i see - sorry that was a naive argument.
: here are some thoughts based on what ppl have discussed so far:
: since every polygon can be triangulated, and every triangle can be
: decomposed into two right triangles, it suffices to prove this
: "measure function" vanishes for all right-angled triangles.
: now take the middle points of each side, along with the right-angled
: vertex. this forms a rectangle; the rest are two similar triangles
: of half of the original size. by repeating this, the origina
|
D**u
发帖数: 204 | 17 This question indeed has an counterexample (that F(P) != 0).
hint: instead of trying to represent F as some integration on P, using the
boundary of P to define F. Make sure that when adding F(P1) and F(P2), something
is "canceled out" on the common boundary of P1 and P2.
region
and P2
【在 D**u 的大作中提到】
: On a 2-dim plane, F is a (real number valued) function on each polygon region
: P. We also know that if P is the union of 2 disjoint polygon regions P1 and P2
: , then
: F(P) = F(P1) + F(P2).
: Question: if for every rectangle D (no need to be parallel to x-y axis) we
: have F(D) = 0, does that imply that F(P) = 0 for every polygon region P?
|
p*****k
发帖数: 318 | 18 DuGu, thanks for the hint. one thing i could think of is to use
contour integral (or in general, line integral), though i have
no idea what could vanish for an arbitrary rectangular contour,
while nonzero for some other particular contour.
(or in terms of the line integral, a field not doing work for
rectangular paths, but not for some other paths) |
D**u
发帖数: 204 | 19 Here is the counterexample (such that F(P) != 0 for some P).
...
Suppose the vertices of polygon P are (in counterclockwise position) z1, ...
, zn, (z_{n+1} := z1). Let theta_i be the angle of vector
z_{i+1} - z_i.
Define g(theta) = sin(3*theta), then we have g(theta) = - g(theta + pi).
We define F(p) = \sum |z_{i+1} - zi| * g(theta_i).
Ease to check the following:
(1) if P is the union of 2 non-intersecting P1 and P2, because the "line integral" on the common boundary of P1 and P2 vanishes, so
【在 p*****k 的大作中提到】
: DuGu, thanks for the hint. one thing i could think of is to use
: contour integral (or in general, line integral), though i have
: no idea what could vanish for an arbitrary rectangular contour,
: while nonzero for some other particular contour.
: (or in terms of the line integral, a field not doing work for
: rectangular paths, but not for some other paths)
|
p*****k
发帖数: 318 | 20 DuGu, very nice construction.
is it possible to extend to curved contours? |
|
|
D**u
发帖数: 204 | 21 I think so. If the curve C is fairly (piecewise) smooth/differentiable, at
each point on the contour let "theta" be the angle of the tangent line.
We can define
F(C) = \integrate g(theta) ds
This question is related to (and different from) a concept called "intrinsic volume" which is also finite additive but not countable additive. The difference is: intrinsic volume is invariant under rigid motions (combinations of rotations and translations), while the construction of F is only invariant under
【在 p*****k 的大作中提到】
: DuGu, very nice construction.
: is it possible to extend to curved contours?
|
p*****k
发帖数: 318 | 22 DuGu, then i'm wondering whether some version of Green's identity
applies here, which could convert this line integral to the surface
integral on P. my intent is to understand what went wrong with the
arguments along that line of thoughts |
D**u
发帖数: 204 | 23 That is a very good question. My guess is that this line integral can not be
converted to a surface integral on P, otherwise we will get F(P)=0 by
applying Riemann sum on the surface integral.
The reason might be that Green's identities generally require the line
integral to integrate some "linear" transformation of the outward pointing
unit normal of line element ds, while g(theta) is not such a "linear"
transformation.
Here is a trivial but interesting modification of the original problem.
Let
【在 p*****k 的大作中提到】
: DuGu, then i'm wondering whether some version of Green's identity
: applies here, which could convert this line integral to the surface
: integral on P. my intent is to understand what went wrong with the
: arguments along that line of thoughts
|
