gugo82 ha scritto:1. Let \( \displaystyle \alpha \in ]0,1] \) .
Prove that there exists \( \displaystyle C\geq 1 \) s.t.
Hölder's elementary inequality:
(
H) \( \displaystyle |x^\alpha -y^\alpha|\leq C\ |x-y|^\alpha \)
holds for all \( \displaystyle x,y\in [0,+\infty[ \) .
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Hints: 1. Since (H) is symmetric with respect to \( \displaystyle x \) and \( \displaystyle y \) (i.e. changing variables' order doesn't change the inequality), without loss of generality you can assume \( \displaystyle x> y \) ; then divide both sides of (H) by \( \displaystyle x^\alpha \) , simplify and introduce the new variable \( \displaystyle t=\tfrac{y}{x} \) .
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 \( \displaystyle \geq 1 \) of the parameter \( \displaystyle C \) such that the ratio of RHside and LHside of (H) is a positive function of \( \displaystyle t \) in \( \displaystyle [0,1] \) with positive minimum.
If \( \displaystyle x=y \) , inequality (
H) holds; hence assume one variable is greater than the other and, w.l.o.g., let \( \displaystyle x>y\geq 0 \) .
Divide both sides of (
H) by \( \displaystyle x \) and introduce the new variable \( \displaystyle t:=\tfrac{y}{x} \in [0,1[ \) : then inequality:
(*) \( \displaystyle \frac{1}{C} (1-t^\alpha) \leq (1-t)^\alpha \) for \( \displaystyle t\in [0,1[ \)
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 \( \displaystyle 1-t^\alpha \) to obtain:
\( \displaystyle \frac{1}{C}\leq \frac{(1-t)^\alpha}{1-t^\alpha} \)
thus (*) will be proved if one can show that the function \( \displaystyle \varphi(t):=\tfrac{(1-t)^\alpha}{1-t^\alpha} \) has positive infimum in \( \displaystyle [0,1[ \) .
The function \( \displaystyle \varphi(t) \) is continuously differentiable in \( \displaystyle ]0,1[ \) and:
\( \displaystyle \varphi^\prime (t)=\frac{\alpha (1-t)^{\alpha -1} (1-t^\alpha) +\alpha t^{\alpha -1} (1-t)^\alpha}{(1-t^\alpha)^2} \)
\( \displaystyle =\frac{\alpha}{(1-t^\alpha)^2}\ \left( \frac{1-t^\alpha}{(1-t)^{1-\alpha}} +\frac{(1-t)^\alpha}{t^{1-\alpha}}\right) \)
\( \displaystyle =\frac{\alpha (t^{1-\alpha} +1)}{t^{1-\alpha}(1-t)^{3-\alpha}} >0 \)
hence \( \displaystyle \varphi(t) \) is strictly increasing in \( \displaystyle [0,1[ \) and \( \displaystyle \inf_{[0,1[} \varphi =\varphi (0)=1>0 \) , therefore the claim.
Moreover \( \displaystyle 1 \) is the best constant in (*), because there is equality when \( \displaystyle t = 0 \) ; therefore \( \displaystyle 1 \) is also the best constant in (
H), with equality when either \( \displaystyle x = 0 < y \) or \( \displaystyle y = 0 < x \) .
gugo82 ha scritto:2. Can \( \displaystyle C=\tfrac{1}{\alpha} \) can be used as a constant in (H)?
Yes, it can be used: actually \( \displaystyle \tfrac{1}{\frac{1}{\alpha}}=\alpha <1=\inf_{[0,1[} \varphi \) , hence \( \displaystyle \alpha < \tfrac{(1-t)^\alpha}{1-t^\alpha} \) for all \( \displaystyle t\in [0,1[ \) .
gugo82 ha scritto:3. Can any constant \( \displaystyle C<\tfrac{1}{\alpha} \) be used in inequality (H) to make it hold for all \( \displaystyle x,y\in [0,+\infty[ \) ?
No, it cannot: in fact \( \displaystyle \tfrac{1}{C} \) has to be less than \( \displaystyle 1 \) to make (*) work in \( \displaystyle [0,1[ \) , hence \( \displaystyle C \) has to be greater than \( \displaystyle 1 \) .
Sono sempre stato, e mi ritengo ancora un dilettante. Cioè una persona che si diletta, che cerca sempre di provare piacere e di regalare il piacere agli altri, che scopre ogni volta quello che fa come se fosse la prima volta. (Freak Antoni)
