[過去ﾛｸﾞ] 現代数学の系譜 工学物理雑談 古典ガロア理論も読む48 (625ﾚｽ)
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496(1): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 2017/12/24(日) 10:31:48.31 ID:Q5UHveEY(9/18) AAS
>>494 つづき
(p. 373) "We omit the proof, because it is rather lengthy,
and one would hope to generalize the theorem by replacing
the rationals by an arbitrary dense set, and possibly to
show that the set of points at which g fails to be
Lipschitzian is a residual set."
NOTE: Sengupta/Lahiri had essentially obtained this result
in 1957 (the points of discontinuity have to form an
F_sigma set, however). See my remark in [13] above.
This result is also proved in Gerald Arthur Heuer,
"A property of functions discontinuous on a dense set",
American Mathematical Monthly 73 #4 (April 1966),
378-379 [MR 34 #2791]. Heuer proves that for each
0 < s <= 1 and for each f:R --> R such that
{x: f is continuous at x} is dense in R and
{x: f is not continuous at x} is dense in R,
the set of points where f does not satisfy a
pointwise Holder condition of order s is the
complement of a first category set (i.e. a co-meager
set). By choosing s < 1, we obtain a stronger version
of Sengupta/Lahiri's result. By intersecting the
co-meager sets for s = 1/2, 1/3, 1/4, ..., we get
a co-meager set G such that, for each x in G, f does
not satisfy a pointwise Holder condition at x for
any positive Holder exponent. (Heuer does not
explicitly state this last result.) A metric space
version of Heuer's result for an arbitrary given
pointwise modulus of continuity condition is essentially
given in: Edward Maurice Beesley, Anthony Perry Morse,
and Donald Chesley Pfaff, "Lipschitzian points",
American Mathematical Monthly 79 #6 (June/July 1972),
603-608 [MR 46 #304; Zbl 239.26004]. See also the last
theorem in Norton [17] below.
つづく
497(5): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 2017/12/24(日) 10:33:54.26 ID:Q5UHveEY(10/18) AAS
>>496 つづき
で、今回の「(a, b) 上でリプシッツ連続である」に関連する部分のみを、さらに抽出すると
[15] Gerald Arthur Heuer先生
THEOREM 4: The function f^2 is Lipschitzian but not
differentiable at the points of the set
{(1/2)*[m - sqrt(d)]: m is an integer
and there exists an integer n such that
d = m^2 - 4n is positive but not a perfect
square} . [This set is dense in the reals.]
THEOREM 5: If g is a function discontinuous at the
rationals and continuous at the irrationals,
then there is a dense uncountable subset
of the reals at each point of which g fails
to satisfy a Lipschitz condition.
かな？
特に、THEOREM 5 変形トマエ函数（Ruler Function）のような、有理数で不連続、無理数で連続なる函数では、
”there is a dense uncountable subset of the reals at each point of which g fails to satisfy a Lipschitz condition.”
だと
だから、（A）”a dense uncountable subset”で、リプシッツ連続は満たさないは、実現できている
では、なぜ、（B）”内点を持たない閉集合の高々可算和”は、実現することができないのか？
[15] Gerald Arthur Heuer先生の（A）と、定理1.7 (422 に書いた定理)の（B）との差！
これを見極めない限り、素人の証明を読んでも仕方が無いと思う
まあ、年末は忙しい
ゆっくりやりましょう（＾＾
以上
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