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How can I prove the following lemma?

Let $1<p<\infty$ 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!

J. Doe
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2 Answers2

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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 $$C_L = \max(C_{L0},C_{L1},C_{L\infty},C_{L1-},C_{L1+}) = C_{L1+} = \color{brown}{\dfrac{1+\varepsilon}{\left(\sqrt[p-1]{1+\varepsilon}-1\right)^{p-1}}}.$$

Yuri Negometyanov
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  • 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 '19 at 02:16
  • @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 '19 at 05:15
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    Yes, I agree with (1) now, and (2) as well. – Calvin Khor Sep 11 '19 at 05:16
  • (I was mistaken earlier, of course $|a|=|x\pm y|$.) But am I reading your $C_L$ correctly? With parameters $\epsilon = 0.01, p=10$, i think $C_L \approx 913.03$ but at $z=2$, $$ |z^p - 1| = 2^{10} - 1 = 1023$$ but $$ \epsilon + C_L |z-1|^p = \epsilon + C_L \approx 913.04 < 1023$$ – Calvin Khor Sep 11 '19 at 05:47
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    @CalvinKhor Was missed degree of the denominators, fixed. Thank you for your work. – Yuri Negometyanov Sep 11 '19 at 06:05
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    Thank you for computing the optimal constant, its much better than the one that I managed to get very quickly – Calvin Khor Sep 11 '19 at 06:11
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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 $|t|<c$, 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.

Calvin Khor
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