推 ERT312 :推一下，如果不用 R over Q 這種超大基底，不知是否 11/09 23:32
We asserted the existence of a Hamel basis for Ｒ over the
field of rationals by appealing to the axiom of choice.
On the other hand, any additive function on Ｒ that is nowhere
continuous is also not Lebesgue measurable, so to prove the
existence of such a function, you need the axiom of choice.
※ 編輯: cgkm 來自: 207.171.15.189 (11/10 04:02)
※ 編輯: cgkm 來自: 207.171.15.189 (11/10 05:54)
※ 引述《gg (GG)》之銘言：
: 標題: [分析] f(x)+f(y)=f(x+y) for all x,y in R.
: 時間: Mon Nov 9 01:20:36 2009
:
: f: R-> R
: f(x)+f(y)=f(x+y) for all x,y in R
:
: If f is continuous at some p in R, it follows that f is continuous on R.
: Then we can conclude that f(x)=cx for all x in R, where c is a constant.
:
: How about the condition that f(x) is discontinuous at each point x of R?
: If possible, please construct an explicit example.
:
: --
: ※ 發信站: 批踢踢實業坊(ptt.cc)
: ◆ From: 114.42.81.224
: 推 zombiea :consider R as Q-vector space with basis {x} 11/09 01:44
: → zombiea :and chose {x} contained in [0,1] 11/09 01:45
: → zombiea :then regard R->R as Q linear transform by 11/09 01:45
: → zombiea :x->1 for all basis x , then this is what you need 11/09 01:46
: 推 cacud :好像泛函的題目@@ 11/09 01:53
: 推 THEJOY :#18YhYIfr 或 搜尋作者THEJOY第二篇 11/09 08:50
Proof (equivalent to THEJOY's approach as modified by me in #1AzwaIPw):
We first show that f is continuous, since for any x and y in Ｒ,
f(x) = f(x - y + x ) + f(y - x ).
0 0
The rest is shown in #1AyZyrWO. □
If f is nowhere continuous, then it is obvious that
f could not be of the form
f(x) = cx, x in Ｒ
because functions of that form are continuous.
Example of a nowhere continuous function (similar to zombiea's approach):
Consider Ｒ as a normed vector space over the field Ｑ
of rational numbers with norm | | (absolute value),
and let B be a (Hamel) basis for this space containing 1 and π.
(We assume the axiom of choice.)
With this norm, the notions of continuity on this vector space
and continuity on the Euclidean real line coincide.
Define the linear operator f:Ｒ→Ｒ by setting
f(π) = 0,
f(b) = b for b in B \ {π}.
By construction, f satisfies
f(x + y) = f(x) + f(y)
for all x and y in Ｒ.
However, if r is a sequence of rational numbers converging
n
to π, then
f(r ) = r → π ≠ 0 = f(π),
n n
so f is not continuous at π, but linear operators
are either continuous everywhere or nowhere continuous.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 75.62.131.87
※ 編輯: cgkm 來自: 75.62.131.87 (11/09 15:07)
※ 編輯: cgkm 來自: 75.62.131.87 (11/09 17:50)