[EX] An elementary inequality for youngsters

[gugo82]

An exercise for young guys, to increase their expertise in Differential Calculus.

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Exercise:

1. Let [tex]$\alpha \in ]0,1]$[/tex].
Prove that there exists [tex]$C\geq 1$[/tex] s.t. Hölder's elementary inequality:

(H) [tex]$|x^\alpha -y^\alpha|\leq C\ |x-y|^\alpha$[/tex]

holds for all [tex]$x,y\in [0,+\infty[$[/tex].

2. Can [tex]$C=\tfrac{1}{\alpha}$[/tex] can be used as a constant in (H)?

3. Can any constant [tex]$C<\tfrac{1}{\alpha}$[/tex] be used in inequality (H) to make it hold for all [tex]$x,y\in [0,+\infty[$[/tex]?

Hints: 1. Since (H) is symmetric with respect to [tex]$x$[/tex] and [tex]$y$[/tex] (i.e. changing variables' order doesn't change the inequality), without loss of generality you can assume [tex]$x> y$[/tex]; then divide both sides of (H) by [tex]$x^\alpha$[/tex], simplify and introduce the new variable [tex]$t=\tfrac{y}{x}$[/tex].
Now you can prove (H) by using Differential Calculus: in fact in order to get (H) it suffices to prove that there exists a value [tex]$\geq 1$[/tex] of the parameter [tex]$C$[/tex] such that the ratio of RHside and LHside of (H) is a positive function of [tex]$t$[/tex] in [tex]$[0,1]$[/tex] with positive minimum.

2 & 3. These are straightforward consequences of 1.

[gugo82]

"gugo82":1. Let [tex]$\alpha \in ]0,1]$[/tex].
Prove that there exists [tex]$C\geq 1$[/tex] s.t. Hölder's elementary inequality:

(H) [tex]$|x^\alpha -y^\alpha|\leq C\ |x-y|^\alpha$[/tex]

holds for all [tex]$x,y\in [0,+\infty[$[/tex].

Hints: 1. Since (H) is symmetric with respect to [tex]$x$[/tex] and [tex]$y$[/tex] (i.e. changing variables' order doesn't change the inequality), without loss of generality you can assume [tex]$x> y$[/tex]; then divide both sides of (H) by [tex]$x^\alpha$[/tex], simplify and introduce the new variable [tex]$t=\tfrac{y}{x}$[/tex].
Now you can prove (H) by using Differential Calculus: in fact in order to get (H) it suffices to prove that there exists a value [tex]$\geq 1$[/tex] of the parameter [tex]$C$[/tex] such that the ratio of RHside and LHside of (H) is a positive function of [tex]$t$[/tex] in [tex]$[0,1]$[/tex] with positive minimum.

If [tex]$x=y$[/tex], inequality (H) holds; hence assume one variable is greater than the other and, w.l.o.g., let [tex]$x>y\geq 0$[/tex].
Divide both sides of (H) by [tex]$x$[/tex] and introduce the new variable [tex]$t:=\tfrac{y}{x} \in [0,1[$[/tex]: then inequality:

(*) [tex]$\frac{1}{C} (1-t^\alpha) \leq (1-t)^\alpha$[/tex] for [tex]$t\in [0,1[$[/tex]

is easily seen to be completely equivalent to (H), therefore it suffices to prove (*) to get the claim.
In order to prove (*), divide both LH and RH sides by [tex]$1-t^\alpha$[/tex] to obtain:

[tex]$\frac{1}{C}\leq \frac{(1-t)^\alpha}{1-t^\alpha}$[/tex]

thus (*) will be proved if one can show that the function [tex]$\varphi(t):=\tfrac{(1-t)^\alpha}{1-t^\alpha}$[/tex] has positive infimum in [tex]$[0,1[$[/tex].
The function [tex]$\varphi(t)$[/tex] is continuously differentiable in [tex]$]0,1[$[/tex] and:

[tex]$\varphi^\prime (t)=\frac{\alpha (1-t)^{\alpha -1} (1-t^\alpha) +\alpha t^{\alpha -1} (1-t)^\alpha}{(1-t^\alpha)^2} $[/tex]
[tex]$=\frac{\alpha}{(1-t^\alpha)^2}\ \left( \frac{1-t^\alpha}{(1-t)^{1-\alpha}} +\frac{(1-t)^\alpha}{t^{1-\alpha}}\right)$[/tex]
[tex]$=\frac{\alpha (t^{1-\alpha} +1)}{t^{1-\alpha}(1-t)^{3-\alpha}} >0$[/tex]

hence [tex]$\varphi(t)$[/tex] is strictly increasing in [tex]$[0,1[$[/tex] and [tex]$\inf_{[0,1[} \varphi =\varphi (0)=1>0$[/tex], therefore the claim.

Moreover [tex]$1$[/tex] is the best constant in (*), because there is equality when [tex]$t = 0$[/tex]; therefore [tex]$1$[/tex] is also the best constant in (H), with equality when either [tex]$x = 0 < y$[/tex] or [tex]$y = 0 < x$[/tex].

"gugo82":2. Can [tex]$C=\tfrac{1}{\alpha}$[/tex] can be used as a constant in (H)?

Yes, it can be used: actually [tex]$\tfrac{1}{\frac{1}{\alpha}}=\alpha <1=\inf_{[0,1[} \varphi$[/tex], hence [tex]$\alpha < \tfrac{(1-t)^\alpha}{1-t^\alpha}$[/tex] for all [tex]$t\in [0,1[$[/tex].

"gugo82":3. Can any constant [tex]$C<\tfrac{1}{\alpha}$[/tex] be used in inequality (H) to make it hold for all [tex]$x,y\in [0,+\infty[$[/tex]?

No, it cannot: in fact [tex]$\tfrac{1}{C}$[/tex] has to be less than [tex]$1$[/tex] to make (*) work in [tex]$[0,1[$[/tex], hence [tex]$C$[/tex] has to be greater than [tex]$1$[/tex].