18/02/25 15:16:00.15 r2nQOARz.net
>>358 つづき
(c) This is not Lipschitz on (0, 1); we prove this in a manner similar to (a).
Define x(n) = 1/√(2πn + π/2) and y(n) = 1/√(2πn) for n = 1, 2, ..., and suppose that f is Lipschitz on (0, 1).
(Note that {x(n)} and {y(n)} are subsets of (0, 1).)
Then, there must exist M > 0 such that
M ? |[f(x(n)) - f(y(n))] / (x(n) - y(n))|
....= |[(2πn + π/2)^(-3/4) - 0] / [1/√(2πn + π/2) - 1/√(2πn)]|
....= √(2πn) / {(2πn + π/2)^(1/4) [√(2πn + π/2) - √(2πn)]}
....= √(2πn) [√(2πn + π/2) + √(2πn)] / {(2πn + π/2)^(1/4) [(2πn + π/2) - (2πn)]}, via conjugates
....= (2/π) √(2πn) [√(2πn + π/2) + √(2πn)] / (2πn + π/2)^(1/4)
....= √(8/π) [√(2πn^2 + πn/2) + n√(2π)] / (2πn + π/2)^(1/4)
....→ ∞ as n→∞. [degree 1 on the numerator, degree 1/4 on the denominator]
Hence, no such M can exist, and so f(x) = x^(3/2) sin(1/x) is not Lipschitz on (0, 1).
(引用終り)