# An inequality involving $|a|^p$: how can I prove this?

How can I prove the following lemma?

Let $$1 and let $$\epsilon >0$$. Then there exists a constant $$C \geq 0$$ (may depend on $$p$$ and $$\epsilon$$) such that for all $$a, b \in \mathbb{C}$$, $$||a+b|^p - |b|^p| \leq \epsilon |b|^p +C|a|^p$$.

I am suffering since I cannot use any homogeneity argument because of the $$|a+b|^p$$ term. I tried to use the fact that $$|t|^p$$ is a convex function for each fixed $$p>1$$, but it does not seem to work, as far as I tried.

Thanks!

• Is $p\in\mathbb{N}$? – C. Brendel Sep 5 at 5:48
• In $\mathbb{R}$. – John Doe Sep 5 at 5:49
• At the moment I am only able to find a proof for $\vert a\vert\ge\vert b\vert$, I find something I'll come back. – C. Brendel Sep 5 at 6:35
• something that may help someone splitting into cases - for $0<|b|$ and $|a|\ll 1$, its like a derivative of $|b|^p$, you get something like $$| |a+b|^p - |b|^p | \le p |a| |b|^{p-1} + o(|a|)$$ where the constant depends on $|b|$, the first term can be dealt with by young's inequality. – Calvin Khor Sep 10 at 12:48
• BTW, my result is optimal (read the comments) – Yuri Negometyanov Sep 13 at 16:55

Let $$\epsilon>0$$ be fixed. The inequality is trivial for $$b=0$$. Else, setting $$t=a/b\in\mathbb C$$, we are asked to prove $$||1+t|^p-1|\le \epsilon +C_{\epsilon,p} |t|^p. \label{*}\tag{*}$$

On the set $$|t|\ge c>0$$, $$c>0$$ to be determined, note that by convexity of $$|t|^p$$, $$\frac{||1+t|^p-1| }{ |t|^p}\le \frac{2^{p-1}+2^{p-1}|t|^p+1}{|t|^p} = \frac{2^{p-1}+1}{|t|^p} + 2^{p-1} \le \frac{2^{p-1}+1}{c^p} + 2^{p-1} =: M_{c,p}.$$ Thus, we have for any $$c$$ $$||1+t|^p-1| \le M_{c,p} |t|^p \le \epsilon + M_{c,p} |t|^p.$$

On the set $$|t|, we have by the Mean Value Inequality and $$\mathbb R^2$$-differentiability of $$|s|^p$$ at $$s=1$$, (writing $$[1,1+t]$$ for the line segment in $$\mathbb R^2\cong \mathbb C$$) $$||1+t|^p - 1| \le \sup_{s\in[1,1+t]} p|s|^{p-1} |t| \le p(1+|t|)^{p-1}|t| \le p(1+|c|)^{p-1}|c|.$$ Now choose $$c=c(\epsilon,p)$$ so small that $$p(1+|c|)^{p-1}|c|< \epsilon,$$ giving $$||1+t|^p - 1| \le \epsilon \le \epsilon + M_{c(\epsilon,p),p} |t|^p$$ We have now proven (\ref{*}), with $$C_{\epsilon,p} = M_{c(\epsilon,p),p}$$.

P.S. $$C_{\epsilon,p}$$ necessarily explodes as $$\epsilon\to 0$$, see here and here.

Let $$x =|a+b|,\quad y = |b|,$$ then the issue inequality takes the form of $$|x^p-y^p| \le \varepsilon y^p + C|x+e^{i\varphi} y|^p,\tag1$$ where the phase $$\varphi$$ depends of the phases of $$(a+b)$$ and $$b.$$

Since for the arbitrary $$u,v\in\mathbb C$$ $$\begin{cases} ||u|-|v|| \le |u+v|\\ ||u|-|v|| \le |u-v|, \end{cases}$$

then the worst case of $$(1)$$ is the case $$|x^p-y^p| \le \varepsilon y^p + C|x - y|^p.\tag2$$

Inequality $$(2)$$ is homogenuis by $$x$$ and $$y,$$ so the least value of $$C$$ should provide the inequality $$|z^p-1| \le \varepsilon + C|z - 1|^p,\tag3$$ where $$z\in[0,\infty).$$

If $$\color{brown}{z=0},$$ then $$C_{L0} =1-\varepsilon.$$

If $$\color{brown}{z=1},$$ then $$C_{L1} =0.$$

If $$\color{brown}{z\to \infty},$$ then $$C_{L\infty} = 1.$$

Denote $$f(C,z) = |z^p-1| - \varepsilon - C|z - 1|^p.\tag4$$

The least value of $$C$$ can be defined from the condition $$\max\limits_{z\in[0,\infty)} f(C,z) = 0.$$

The inner maxima can be achived only if $$f'_z(C,z)=0.$$

If $$\color{brown}{z\in(0,1)}$$ then $$f(C,z) = 1-z^p-\varepsilon - C(1-z)^p,$$

$$\begin{cases} -pz^{p-1}+Cp(1-z)^{p-1} = 0\\ 1-z^p-\varepsilon - C(1-z)^p = 0, \end{cases}$$

$$\begin{cases} \dfrac z{1-z}=\sqrt[p-1]C,\quad z = \dfrac{\sqrt[p-1]C}{\sqrt[p-1]C+1}\\ 1 - \varepsilon = \left(\dfrac{\sqrt[p-1]C}{\sqrt[p-1]C+1}\right)^p + C\left(\dfrac{1}{\sqrt[p-1]C+1}\right)^p, \end{cases}$$

$$1 - \varepsilon = \dfrac{C\left(\sqrt[p-1]C+1\right)}{\left(\sqrt[p-1]C+1\right)^p} = \dfrac{C}{\left(\sqrt[p-1]C+1\right)^{p-1}} = z^{p-1},\quad z=\sqrt[p-1]{1-\varepsilon},$$ $$C_{L1-}= \left(\dfrac z{1-z}\right)^{p-1} = \dfrac{1-\varepsilon}{\left(1-\sqrt[p-1]{1-\varepsilon}\right)^{p-1}}.$$

If $$\color{brown}{z\in(1,\infty)}$$ then $$f(C,z) = z^p-1-\varepsilon - C(z-1)^p,$$

$$\begin{cases} pz^{p-1}-Cp(z-1)^{p-1} = 0\\ z^p-1-\varepsilon - C(z-1)^p = 0, \end{cases}$$

$$\begin{cases} \dfrac z{z-1}=\sqrt[p-1]C,\quad z = \dfrac{\sqrt[p-1]C}{\sqrt[p-1]C-1}\\ 1 + \varepsilon = \left(\dfrac{\sqrt[p-1]C}{\sqrt[p-1]C-1}\right)^p - C\left(\dfrac{1}{\sqrt[p-1]C-1}\right)^p, \end{cases}$$

$$1 + \varepsilon = \dfrac{C\left(\sqrt[p-1]C-1\right)}{\left(\sqrt[p-1]C-1\right)^p} = \dfrac{C}{\left(\sqrt[p-1]C-1\right)^{p-1}} = z^{p-1},\quad z=\sqrt[p-1]{1+\varepsilon},$$

$$C_{L1+}= \left(\dfrac z{z-1}\right)^{p-1} = \dfrac{1+\varepsilon}{\left(\sqrt[p-1]{1+\varepsilon}-1\right)^{p-1}}.$$

Since $$\dfrac{C_{L1-}}{C_{L1+}} = \dfrac{1-\varepsilon}{1+\varepsilon}\left(\dfrac{\sqrt[p-1]{1+\varepsilon}-1}{1-\sqrt[p-1]{1-\varepsilon}}\right)^{p-1} = \left(\dfrac{\sqrt[p-1]{1-\varepsilon}}{\sqrt[p-1]{1+\varepsilon}}\dfrac{\sqrt[p-1]{1+\varepsilon}-1}{1-\sqrt[p-1]{1-\varepsilon}}\right)^{p-1}$$ $$= \left(\dfrac{\sqrt[p-1]{1-\varepsilon^2}-\sqrt[p-1]{1-\varepsilon}}{\sqrt[p-1]{1+\varepsilon}-\sqrt[p-1]{1-\varepsilon^2}}\right)^{p-1} = \left(1+\dfrac{2\ \sqrt[p-1]{1-\varepsilon^2}-\sqrt[p-1]{1-\varepsilon}-\sqrt[p-1]{1+\varepsilon}}{\sqrt[p-1]{1+\varepsilon}-\sqrt[p-1]{1-\varepsilon^2}}\right)^{p-1} \le1,$$

then the least value of the constant $$C$$ is $$C_L = \max(C_{L0},C_{L1},C_{L\infty},C_{L1-},C_{L1+}) = C_{L1+} = \color{brown}{\dfrac{1+\varepsilon}{\left(\sqrt[p-1]{1+\varepsilon}-1\right)^{p-1}}}.$$

• I don't follow the initial reduction step, what is the sign of $a+b,b\in\mathbb C$? Even if they are real valued, its not always true that $|a| = |x\pm y|$. The right hand side should be $\epsilon |b|^p + C |c-b|^p$ where $c = a+b \in \mathbb C$. I suppose you want to say that you will instead prove the stronger inequality $$| |c|^p - |b|^p | \le \epsilon |b|^p + C ||c|-|b||^p ?$$ – Calvin Khor Sep 11 at 2:16
• @CalvinKhor Thank you for the comment, $(1)$ is fixed for the complex case. At the same time, $x$ and $y$ are real positive, so my formulalion of $(2)$ is right. – Yuri Negometyanov Sep 11 at 5:15
• Yes, I agree with (1) now, and (2) as well. – Calvin Khor Sep 11 at 5:16
• @CalvinKhor Was missed degree of the denominators, fixed. Thank you for your work. – Yuri Negometyanov Sep 11 at 6:05
• Thank you for computing the optimal constant, its much better than the one that I managed to get very quickly – Calvin Khor Sep 11 at 6:11