# A certain inequality bound

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.

• Have you more detailed attempts ?
– EDX
Commented Jun 12 at 11:00
• 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. Commented Jun 12 at 11:05
• Would you mind sharing where you saw that estimate, in case the context could be useful? Commented Jun 12 at 17:17

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.

• 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 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.