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Question Number 75041 by ~blr237~ last updated on 06/Dec/19

1) Show that for a \in] 01] the function f_{a} : R_{+} \to R defined by f_{a} (x) = x^{a}

is a - holder function in other way there exist K > 0 such as \forall x, y > 0

∣ f_{a} (x) - f_{a} (y) ∣⩽ K ∣ x - y ∣^{a}

Commented by ~blr237~ last updated on 06/Dec/19

for a \in] 0, 1 [

Answered by mind is power last updated on 06/Dec/19

error my previous post i put inquality in wrong why

i found this new approch

x^{a} - y^{a} ⩽ {(x - y)}^{a} \dots ?

\Leftrightarrow 1 - {(\frac{y}{x})}^{a} ⩽ {(1 - \frac{y}{x})}^{a}, < 0 < y < x

t = \frac{y}{x} \in] 0, 1 [

1 - t^{a} ⩽ {(1 - t)}^{a} \Leftrightarrow {(1 - t)}^{a} + t^{a} ⩾ 1

let f_{t} (a) = {(1 - t)}^{a} + t^{a} = e^{aln (1 - t)} + e^{aln (t)}

f^{'} (a) = \ln (1 - t) e^{aln (1 - t)} + \ln (t) e^{aln (t)} < 0, sincet \in] 0, 1 [

\Rightarrow f (a) ⩾ f (1) = 1 - t + t = 1

\Rightarrow 1 ⩽ t^{a} + {(1 - t)}^{a}

\Rightarrow x^{a} - y^{a} ⩽∣ x - y ∣^{a}

f_{a} is holder k = 1