現代数学の系譜工学物理雑談古典ガロア理論も読む49

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40: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2017/12/28(木) 23:41:49.82 ID:IsA0R4yK

>>30
沢山のレスがありがとう
まあ、ゆっくりやろう

まだ、疑問に思っているのは
下記のDifferentiability of the Ruler Functionの記述と貴方の定理との整合性だ

http://mathforum.org/kb/message.jspa?messageID=5432910 （>>35より）
Topic: Differentiability of the Ruler Function Dave L. Renfro Posted: Dec 13, 2006 Replies: 3 Last Post: Jan 10, 2007
(抜粋)
The ruler function f is defined by f(x) = 0 if x is
irrational, f(0) = 1, and f(x) = 1/q if x = p/q
where p and q are relatively prime integers with q > 0.

Using ruler-like functions that "damp-out" quicker
than any power of f gives behavior that one would
expect from the above.

Let w:Z+ --> Z+ be an increasing function that
eventually majorizes every power function. Define
f_w(x) = 0 for x irrational, f_w(0) = 1, and
f_w(p/q) = 1/w(q) where p and q are relatively
prime integers.

** f_w is differentiable on a set whose complement
has Hausdorff dimension zero. Jurek [4] (pp. 24-25)

Interesting, each of the sets of points where these
functions fail to be differentiable is large in the
sense of Baire category.

THEOREM: Let g be continuous and discontinuous on sets
of points that are each dense in the reals.
Then g fails to have a derivative on a
co-meager (residual) set of points. In fact,
g fails to satisfy a pointwise Lipschitz
condition, a pointwise Holder condition,
or even any specified pointwise modulus of
continuity condition on a co-meager set.

(Each co-meager set has c points in every interval.)

つづく

http://rio2016.5ch.net/test/read.cgi/math/1514376850/40

41: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2017/12/28(木) 23:44:24.28 ID:IsA0R4yK

>>40 つづく

[13] Gerald Arthur Heuer, "Functions continuous at irrationals and discontinuous at rationals", abstract of talk given 2 November 1963 at the annual fall meeting of the Minnesota Section of the MAA, American Mathematical Monthly 71 #3 (March 1964), 349.

The complete text of the abstract follows, with minor editing changes to accommodate ASCII format.

Earlier results of Porter, Fort, and others suggest additional questions about the functions in the title. Differentiability and Lipschitz conditions are considered. Special attention ispaid to the ruler function (f) and its powers.
Sample results:
THEOREM:
If 0 < r < 2, f^r is nowhere Lipschitzian; f^2 is nowhere differentiable, but is Lipschitzian on a dense subset of the reals.
THEOREM:
If r > 0, f^r is continuous but not Lipschitzian at every Liouville number;
if r > 2, f^r is differentiable at every algebraic irrational.
THEOREM:
If g is continuous at the irrationals and not continuous at the rationals, then there exists a dense uncountable subset of the reals at each point of which g fails to satisfy a Lipschitz condition.

REMARK BY RENFRO:
The last theorem follows from the following stronger and more general result.
Let f:R --> R be such that the sets of points at which f is continuous and discontinuous are each dense in R.
Let E be the set of points at which f is continuous and where at least one of the four Dini derivates of f is infinite.
Then E is co-meager in R (i.e. the complement of a first category set).
This was proved in H. M. Sengupta and B. K. Lahiri, "A note on derivatives of a function",
Bulletin of the Calcutta Mathematical Society 49 (1957), 189-191 [MR 20 #5257; Zbl 85.04502]. See also my note in item [15] below.
(引用終り)

つづく

http://rio2016.5ch.net/test/read.cgi/math/1514376850/41

245: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2018/01/07(日) 20:46:08.89 ID:2l42E8SE

>>238
おっちゃん、どうも、スレ主です。

>いっておくけど、系 1.8 の結果は
>有理数の点で不連続, 無理数の点で一様連続となるf : R → R は存在しない
>まで拡張出来る。

ああ、下記だな”g fails to satisfy・・even any specified pointwise modulus of continuity condition on a co-meager set.”＆
”Let E be the set of points at which f is continuous and where at least one of the four Dini derivates of f is infinite.
Then E is co-meager in R (i.e. the complement of a first category set).”

繰返すが、
”any specified pointwise modulus of continuity condition” ＆ ”at least one of the four Dini derivates of f is infinite”
だから（特に後者Dini微分）、どこかの無理数の点で一様連続も破綻するだろうな

（>>40より）http://mathforum.org/kb/message.jspa?messageID=5432910
THEOREM: Let g be continuous and discontinuous on sets of points that are each dense in the reals.
Then g fails to have a derivative on a co-meager (residual) set of points.
In fact, g fails to satisfy a pointwise Lipschitz condition, a pointwise Holder condition, or even any specified pointwise modulus of continuity condition on a co-meager set.

(Each co-meager set has c points in every interval.)

（>>41より）
REMARK BY RENFRO:
The last theorem follows from the following stronger and more general result.
Let f:R --> R be such that the sets of points at which f is continuous and discontinuous are each dense in R.
Let E be the set of points at which f is continuous and where at least one of the four Dini derivates of f is infinite.
Then E is co-meager in R (i.e. the complement of a first category set).
This was proved in H. M. Sengupta and B. K. Lahiri, "A note on derivatives of a function",
Bulletin of the Calcutta Mathematical Society 49 (1957), 189-191 [MR 20 #5257; Zbl 85.04502]. See also my note in item [15] below.
(引用終り)

http://rio2016.5ch.net/test/read.cgi/math/1514376850/245

366: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2018/01/10(水) 21:42:07.86 ID:xixJS48Q

>>364 追加

ちょっと思いついたので、悪いが、忘れないうちに下記を書いておく

（>>40より）
http://mathforum.org/kb/message.jspa?messageID=5432910 （>>35より）
Topic: Differentiability of the Ruler Function Dave L. Renfro Posted: Dec 13, 2006 Replies: 3 Last Post: Jan 10, 2007
より
The modefied ruler function f is defined by
f(x) = 0 if x is irrational,
f(0) = 1, and
f(x) = 1/w(q) if x = p/q ∈Q
where p and q are relatively prime integers with q > 0.

ここに
w(q)：an increasing function that eventually majorizes every power function.
（w(q)は、どんなｐの冪より早く増大する関数
　https://kbeanland.files.wordpress.com/2010/01/beanlandrobstevensonmonthly.pdf
　Modifications of Thomae’s function and differentiability, (with James Roberts and Craig Stevenson) Amer. Math. Monthly, 116 (2009), no. 6, 531-535.
　などではP532で、” (e.g., ai = 1/i^(i^i) )”などと記されている。qで書けば、= 1/q^(q^q)だ）

簡単のために、区間[0, 1]を考える。（同じことを、区間[n, n+1] (ｎは整数)で考えれば、実数R全体に展開できる）

このような、場合、上記数学者のRenfroさんや、Robertsさんたちは、”Qで不連続、リュービル数（超越数）で微分不可（リプシッツ連続でもない）だが、それ以外の無理数では、微分可だ”という

つづく

http://rio2016.5ch.net/test/read.cgi/math/1514376850/366

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