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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
$$\left.\mathrm{1}\left.\right)\left.\:\mathrm{Show}\:\mathrm{that}\:\:\mathrm{for}\:\mathrm{a}\in\right]\mathrm{01}\right]\mathrm{the}\:\mathrm{function}\:\:\mathrm{f}_{\mathrm{a}} \::\mathbb{R}_{+} \rightarrow\mathbb{R}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{f}_{\mathrm{a}} \left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{a}} \: \\ $$$$\mathrm{is}\:\:\:\mathrm{a}−\mathrm{holder}\:\mathrm{function}\:\:\mathrm{in}\:\mathrm{other}\:\mathrm{way}\:\:\mathrm{there}\:\mathrm{exist}\:\:\mathrm{K}>\mathrm{0}\:\mathrm{such}\:\mathrm{as}\:\forall\:\mathrm{x},\mathrm{y}>\mathrm{0}\: \\ $$$$\mid\mathrm{f}_{\mathrm{a}} \left(\mathrm{x}\right)−\mathrm{f}_{\mathrm{a}} \left(\mathrm{y}\right)\mid\leqslant\mathrm{K}\mid\mathrm{x}−\mathrm{y}\mid^{\mathrm{a}} \:\: \\ $$$$ \\ $$
Commented by ~blr237~ last updated on 06/Dec/19
for  a∈]0,1[
$$\left.\mathrm{for}\:\:\mathrm{a}\in\right]\mathrm{0},\mathrm{1}\left[\:\right. \\ $$
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
$$\mathrm{error}\:\mathrm{my}\:\mathrm{previous}\:\mathrm{post}\:\mathrm{i}\:\mathrm{put}\:\mathrm{inquality}\:\mathrm{in}\:\mathrm{wrong}\:\mathrm{why} \\ $$$$\mathrm{i}\:\mathrm{found}\:\mathrm{this}\:\mathrm{new}\:\mathrm{approch} \\ $$$$\mathrm{x}^{\mathrm{a}} −\mathrm{y}^{\mathrm{a}} \leqslant\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{a}} …? \\ $$$$\Leftrightarrow\mathrm{1}−\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{a}} \leqslant\left(\mathrm{1}−\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{a}} ,<\mathrm{0}<\mathrm{y}<\mathrm{x} \\ $$$$\left.\mathrm{t}=\frac{\mathrm{y}}{\mathrm{x}}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\mathrm{1}−\mathrm{t}^{\mathrm{a}} \leqslant\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{a}} \Leftrightarrow\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{a}} +\mathrm{t}^{\mathrm{a}} \geqslant\mathrm{1} \\ $$$$\mathrm{let}\:\mathrm{f}_{\mathrm{t}} \left(\mathrm{a}\right)=\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{a}} +\mathrm{t}^{\mathrm{a}} =\mathrm{e}^{\mathrm{aln}\left(\mathrm{1}−\mathrm{t}\right)} +\mathrm{e}^{\mathrm{aln}\left(\mathrm{t}\right)} \\ $$$$\left.\mathrm{f}'\left(\mathrm{a}\right)=\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)\mathrm{e}^{\mathrm{aln}\left(\mathrm{1}−\mathrm{t}\right)} +\mathrm{ln}\left(\mathrm{t}\right)\mathrm{e}^{\mathrm{aln}\left(\mathrm{t}\right)} <\mathrm{0},\mathrm{sincet}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{a}\right)\geqslant\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{t}+\mathrm{t}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}\leqslant\mathrm{t}^{\mathrm{a}} +\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{a}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{a}} −\mathrm{y}^{\mathrm{a}} \leqslant\mid\mathrm{x}−\mathrm{y}\mid^{\mathrm{a}} \\ $$$$\mathrm{f}_{\mathrm{a}} \:\mathrm{is}\:\mathrm{holder}\:\mathrm{k}=\mathrm{1} \\ $$$$ \\ $$

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