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

1) Show that for a∈]01]the function f_a :R_+ →R defined by f_a (x)=x^a is a−holder function in other way there exist K>0 such as ∀ x,y>0 ∣f_a (x)−f_a (y)∣≤K∣x−y∣^a

1)Showthatfora∈]01]thefunctionfa:R+→Rdefinedbyfa(x)=xaisa−holderfunctioninotherwaythereexistK>0suchas∀x,y>0∣fa(x)−fa(y)∣⩽K∣x−y∣a

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

for a∈]0,1[

fora∈]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 ...? ⇔1−((y/x))^a ≤(1−(y/x))^a ,<0<y<x t=(y/x)∈]0,1[ 1−t^a ≤(1−t)^a ⇔(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∈]0,1[ ⇒f(a)≥f(1)=1−t+t=1 ⇒1≤t^a +(1−t)^a ⇒x^a −y^a ≤∣x−y∣^a f_a is holder k=1

errormypreviouspostiputinqualityinwrongwhyifoundthisnewapprochxa−ya⩽(x−y)a...?⇔1−(yx)a⩽(1−yx)a,<0<y<xt=yx∈]0,1[1−ta⩽(1−t)a⇔(1−t)a+ta⩾1letft(a)=(1−t)a+ta=ealn(1−t)+ealn(t)f′(a)=ln(1−t)ealn(1−t)+ln(t)ealn(t)<0,sincet∈]0,1[⇒f(a)⩾f(1)=1−t+t=1⇒1⩽ta+(1−t)a⇒xa−ya⩽∣x−y∣afaisholderk=1

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