4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Is $p\in\mathbb{N}$? $\endgroup$ – C. Brendel Sep 5 at 5:48
  • $\begingroup$ In $\mathbb{R}$. $\endgroup$ – John Doe Sep 5 at 5:49
  • $\begingroup$ 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. $\endgroup$ – C. Brendel Sep 5 at 6:35
  • $\begingroup$ 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. $\endgroup$ – Calvin Khor Sep 10 at 12:48
  • 1
    $\begingroup$ BTW, my result is optimal (read the comments) $\endgroup$ – Yuri Negometyanov Sep 13 at 16:55
1
$\begingroup$

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.

$\endgroup$
2
+50
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $$ | |c|^p - |b|^p | \le \epsilon |b|^p + C ||c|-|b||^p ?$$ $\endgroup$ – Calvin Khor Sep 11 at 2:16
  • $\begingroup$ @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. $\endgroup$ – Yuri Negometyanov Sep 11 at 5:15
  • 1
    $\begingroup$ Yes, I agree with (1) now, and (2) as well. $\endgroup$ – Calvin Khor Sep 11 at 5:16
  • 1
    $\begingroup$ @CalvinKhor Was missed degree of the denominators, fixed. Thank you for your work. $\endgroup$ – Yuri Negometyanov Sep 11 at 6:05
  • 1
    $\begingroup$ Thank you for computing the optimal constant, its much better than the one that I managed to get very quickly $\endgroup$ – Calvin Khor Sep 11 at 6:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.