Why does the following estimate hold for any $p \in [1,\infty)$, and $\delta \in (0,1)$, $$ \big ||a-b|^p - |a|^p - |b|^p \big | \leq \delta |a|^p + \frac{C_p}{\delta^p} |b|^p $$ If we use convexity of $|x|^p$ and use something like Jensen's inequality, there won't be delta terms in front of $|a|^{p}$. Thank you.

  • $\begingroup$ Have you more detailed attempts ? $\endgroup$
    – EDX
    Commented Jun 12 at 11:00
  • $\begingroup$ unfortunately..no others attempts came to my mind.. appeareance of $\delta^{p}$ in numerator and denominator also suggest to use young inquality result that is seen wikipedia, which also seems to fail because there is just p. $\endgroup$ Commented Jun 12 at 11:05
  • $\begingroup$ Would you mind sharing where you saw that estimate, in case the context could be useful? $\endgroup$
    – Bruno B
    Commented Jun 12 at 17:17

2 Answers 2


In this post it has been shown - in the notation of that post - that $$||x+y|^t-|x|^t-|y|^t|\leq C(|x|^{t}|y/x|+|x/y||y|^{t})$$

where $t>1$ and $x,y$ are real numbers and $C$ is only dependent on $t$.

Now, for $a>c$, let $x=a$ and $y=c$ giving $0<|y/x|=|c/a|=\delta^t<1$, giving

$$||a+c|^t-|a|^t-|c|^t|\leq C(\delta^t|a|^{t}+\frac{1}{\delta^t}|c|^{t})$$

Now let $b = -c$ and $t=p$ to achieve

$$||a-b|^p-|a|^p-|b|^p|\leq C(\delta^p|a|^{p}+\frac{1}{\delta^p}|b|^{p})$$

I reckon that should be the answer. In particular, in your question there should be $\delta^p|a|^{p}$, not $\delta|a|^{p} $. Since for symmetry reasons, exchanging $a \leftrightarrow b$ does not change the LHS, so in the RHS the constant cannot be pinned to only one variable, and also the $\delta^p$'s should appear on both variables.

  • $\begingroup$ OP's inequality could still be true even with an asymmetry, in fact the way to make it symmetrical again would be to simply use a $\min$ on the RHS between the current RHS and the one where $a$ and $b$ are swapped $\endgroup$
    – Bruno B
    Commented Jun 12 at 17:12

I was inspired by zhw's answer to this post: https://math.stackexchange.com/q/2853822/1104384.\ This is a partial answer because I wasn't yet able to turn my constant $C_{p,\delta}$ into the desired $C_p/\delta^p$, but maybe someone else can use this as a first step to proving the expected inequality.

Suppose $a \neq 0$ (the inequality will then be true for $a = 0$ if it's true otherwise).

Then $$\big ||a-b|^p - |a|^p - |b|^p \big | \leq \delta |a|^p + C_{p,\delta} |b|^p$$ is true iff the following is true, where $c := b/a$ : $$\big ||1-c|^p - 1 - |c|^p \big | \leq \delta + C_{p,\delta} |c|^p$$ Now, consider the functions $$f : c \in \mathbb{R} \,\,\longmapsto\,\, |1-c|^p - 1 - |c|^p\\ g : c \in \mathbb{R}\setminus\{0\} \,\,\longmapsto\,\, \frac{f(c)}{|c|^p} = \frac{|1-c|^p - 1 - |c|^p}{|c|^p} = \left|\frac{1}{c}-1\right|^p - \frac{1}{|c|^p} - 1 = f\left(\frac{1}{c}\right)$$ We have that $f(c) \to 0$ as $c \to 0$, thus there exists $\eta > 0$ such that: $$|c| \leq \eta \quad\Longrightarrow\quad |f(c)| \leq \delta$$ Moreover, $g$ is continuous on the closed set $\{|c| \geq \eta\}$ and $g(c)$ tends to $0$ as $c \to \infty$, hence one can show that $g$ is bounded on $\{|c| \geq \eta\}$ by a constant $C_{p,\delta} \geq 0$, hence $|f(c)| \leq C_{p,\delta} |c|^p$ if $|c| \geq \eta$.

This yields: $$\forall c \in \mathbb{R},\quad \big ||1-c|^p - 1 - |c|^p \big |\,\, =\,\, |f(c)| \,\,\leq\,\, \delta + C_{p,\delta} |c|^p$$ and we are done with this partial answer by our previous observations.


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