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I requiere some inequalities generalization. I don't know whether they are true or not. Can you help me ?

Here we're talking about $L^p$ spaces with $p > 1$.

I know that on the real line :

$$ ||x|-|y|| \leq | x-y | \leq |x|+|y| $$ equivalently :

$$ ||x|-|y|| \leq | x+y | \leq |x|+|y|$$

Now i'm trying to find similar inequalities in Lebesgues spaces.

I already found that one :

$$(|x + y|)^p \leq 2^{p-1} (|x|^p + |y|^p)$$ thanks to Jensen ineqality.

I also know Minkowski inequality's telling me :

$$ \|f + g\|_{L^p} \leq \|f\|_{L^p} + \|g\|_{L^p}$$

Now I'm searching for something on the other boundary. Meaning, as my friends told me should be true :

$$ |\|f\|_{L^p} - \|g\|_{L^p} | \leq \|f-g\|_{L^p}$$ and equivalently :

$$ |\|f\|_{L^p} - \|g\|_{L^p} | \leq \|f+g\|_{L^p}$$

i would also like to find something like this :

$$\lambda |(|x|^p - |y|^p)| \leq (|x + y|)^p $$

Do you know if something like those 2 inequalities exists, and if yes, how do you prove them ?

thanks !

Pink Panther
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Marine Galantin
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2 Answers2

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In any normed space $$||x||=||(x-y)+y||\le||x-y||+||y||,$$hence $$||x||-||y||\le||x-y||.$$Similarly (or "hence, swapping $x$ and $y$"), $$||y||-||x||\le||x-y||.$$ If $a,b\in\Bbb R$ then $|a-b|=\max(a-b,b-a)$, so $$\big|\,||x||-||y||\,\big|\le||x-y||.$$

David C. Ullrich
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  • could you please also take a look at the other part, which is related to what you said in a way ? or do you agree with our fellow @Calvin Khor ? – Marine Galantin Jun 17 '19 at 17:11
  • @MarineGalantin Looks like the other answer takes care of the other part... – David C. Ullrich Jun 17 '19 at 17:14
  • yeah ok i agree with you. I just don't know to whom give the ok answer. I already upvoted the other post, so i think since you did the major part of the answer you deserve the most. – Marine Galantin Jun 17 '19 at 17:16
  • @MarineGalantin Don't worry about that, accept whichever you like. If you can't decide, note he could use the points more than me... – David C. Ullrich Jun 17 '19 at 17:19
  • If instead of looking at the norm $$ || \cdot ||{L^p} $$ we look at $$ || \cdot ||{L^p}^p $$ you agree that the inequalities we talked about aren't true anymore if you try to for instance just distribute the power ? For instance, simply, the Minkowski inequality won't be true? And you can't majorate an expression like : $$ ||f||{L^p}^p - ||g||{L^p}^p $$ as I wanted, because of Calvin's reference ? The best you could do it taking the exponents on both side of the inequation? which would give $$ ( \big|, ||f||{L^p} - ||g||{L^p} ,\big| )^p \le||f-g||_{L^p}^p $$ – Marine Galantin Jun 17 '19 at 17:27
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    @MarineGalantin Right - Calvin's reference shows it fails for the $p$th power of the norm. – David C. Ullrich Jun 17 '19 at 18:04
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The result in the quote box is true by exactly the same proof as in $\mathbb R$.

The second part of your question is essentially a duplicate of this older question of mine, Can I expect $|x|^s - |y|^s \leq C|x-y|^s$ for $s>1$? (with related discussion) where a counterexample (valid already in dimension 1) shows that $||x|^p - |y|^p| \le C |x-y|^p$ is not possible for $p>1$. in 1D, the counterexample is obtained by choosing $$x=x_0,y=x_0 + t, \quad x_0 > 0,\quad 0<t\ll 1.$$

Essentially, this is because $|x|^p$ is $C^1$ when $p>1$ so we have the (local) estimate from Mean Value Theorem, $$ |x+h|^p - |x|^p \approx p\operatorname{sgn}(x)|x|^{p-1} h.$$ That is, the leading error term is linear in $h$, and its not possible to obtain a bound with a higher order error term than $|h|$.

Calvin Khor
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