14.3 Lipschitz and Hölder functions[2DR]
[162] Let \(A⊂ ℝ\). A function \(f:A→ℝ \) is said Lipschitz continuous if there exists \(L{\gt}0\) such that \(∀ x,y∈ A\),
A function \(f:A→ℝ \) is said Hölder continuous if \(L{\gt}0\) and \(𝛼∈ (0,1]\) exist such that \(∀ x,y∈ A\),
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[163]Prerequisites:1.Show that the Lipschitz functions, as well as Hölder functions, are uniformly continuous What can be said about their continuity modulus?
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[164]Let \(I⊂ ℝ\) be an open interval. Let \(f : I → ℝ\) be differentiable. Show that \(f'\) is bounded on \(I\), if and only if \(f\) is Lipschitz continuous.
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[165]Let \(I⊂ ℝ\) interval. Let \(f : I → ℝ\) such that there exists \(α {\gt} 1\) such that \(∀ x, y, |f (x) − f (y)| ≤ |x − y|^α\) (i.e. \(f\) is Hölder continuous of order \(𝛼{\gt}1\)): Show that f is constant.
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[166]Let be given \(f : [a, b] →ℝ\) and a decomposition of \([a, b]\) into intervals \(I_ 1 = [a, t_ 1 ], I_ 2 = [t_ 1 , t_ 2 ], \ldots , I_ n = [t_{n-1} , b]\) such that the restriction of \(f\) on each \(I_ k\) is Lipschitz of constant \(C\). Show that \(f\) is Lipschitz of constant \(C\).
Similarly for Hölder functions.
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[167]Let \(f : [a, b] →ℝ\) Hölder with exponent \(𝛼 ≤ 1\). Show that f is Hölderian with exponent \(𝛽\) for every \(𝛽 {\lt} 𝛼\).
Note that this is not technically true for \(f:ℝ→ℝ\).
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[169] Build \(f : [0, 1] →ℝ\) that is continuous but not Hölder continuous. Hidden solution: [UNACCESSIBLE UUID ’16B’][UNACCESSIBLE UUID ’16C’]
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[16D] A linear function \(f:ℝ^ n→ℝ^ k\) is Lipschitz.
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[16F]For each of the following functions, say if it is continuous, uniformly continuous, Hölder (and with which exponent), or Lipschitz.
\(f:(0,1)→ ℝ\), \(f(x)=\sin (1/x)\).
\(f:(0,1)→ ℝ\), \(f(x)=x^{1/x}\).
\(f:(1,∞)→ ℝ\), \(f(x)=\sin (x^ 2)/x\)
\(f:[-1,1]→ℝ\), \(f(x)=|x|^𝛽\) with \(𝛽{\gt}0\).
\(f:(0,∞)→ ℝ\), \(f(x)=\sin (x^𝛽)\) with \(𝛽{\gt}0\).
Hidden solution: [UNACCESSIBLE UUID ’16H’]
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[16J] Given \(L∈(0,1)\) if \(f:ℝ→ℝ\) satisfies
\[ |f(x)-f(y)|≤ L|x-y|\quad ∀ x,y∈ℝ \]Then there is only one ”fixed point” that is a point \(x\) for which \(f(x)=x\).
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[16K] Find a function \(f:ℝ→ℝ\) such that
\[ |f(x)-f(y)|{\lt}|x-y|\quad ∀ x,y∈ℝ \]but for which there is no ”fixed point” (that is a point \(x\) for which \(f(x)=x\)=.
Hidden solution: [UNACCESSIBLE UUID ’16M’]