Proof of the inequality $||a+b|^p - |a|^p| \leq \epsilon |a|^p + C_{\epsilon} |b|^p$

Question

In this paper, the following theorem is used implicitly (see Example (a) on page 488):

Let $0<p<\infty$. For every $\epsilon > 0$, there exists some $C_{\epsilon} > 0$ such that $\forall a,b \in \mathbb{C}$: $$ \left||a+b|^p - |a|^p \right| \leq \epsilon |a|^p + C_{\epsilon} |b|^p $$

Is this a well-known result? I would like to know its proof, and its name if it has one. Can someone point me towards a reference to this result which includes a proof, or give me a proof outline?

I've tried using Jensen's Inequality for the case $p>1, a,b>0$, but it doesn't seem to work out.

Answer 1

Define $k=\dfrac{b}{a}$. This theorem is equivalent to show that $$\left||1+k|^p-1\right|\le\epsilon+C_\epsilon|k|^p$$First let $|1+k|\ge 1$, then we must have$$|1+k|^p-1\le\epsilon+C_\epsilon|k|^p$$or equivalently$$\dfrac{|1+k|^p-1-\epsilon}{|k|^p}\le C_\epsilon$$if $|1+k|^p\le1+\epsilon$ the theorem holds with any $C_\epsilon>0$ then assume that $|1+k|^p>1+\epsilon$. By triangle inequality we have $$(1+|k|)^p\ge|1+k|^p>1+\epsilon$$which yields to $$|k|>(1+\epsilon)^{\frac{1}{p}}-1$$or$$|\dfrac{1}{k}|<\dfrac{1}{(1+\epsilon)^{\frac{1}{p}}-1}$$therefore $$\dfrac{|1+k|^p-1-\epsilon}{|k|^p}<(1+|\dfrac{1}{k}|)^p<\dfrac{1+\epsilon}{((1+\epsilon)^{\frac{1}{p}}-1)^p}$$then by choosing $C_\epsilon=\dfrac{1+\epsilon}{((1+\epsilon)^{\frac{1}{p}}-1)^p}$ we complete our proof in this case.

The other case is where $|1+k|<1$. Here we need to show that $$\dfrac{1-|1+k|^p-\epsilon}{|k|^p}\le C_\epsilon$$ for $|1+k|^p\ge 1-\epsilon$ this automatically holds and for $|1+k|^p<1-\epsilon$ we have $$|1+k|<(1-\epsilon)^{\dfrac{1}{p}}$$this is inside a circle of radius $(1-\epsilon)^{\dfrac{1}{p}}$ centered at $-1$ which geometrically means that $$|k|>1-(1-\epsilon)^{\dfrac{1}{p}}$$. Here taking $C_\epsilon=\dfrac{1}{\left(1-(1-\epsilon)^{\dfrac{1}{p}}\right)^p}$ fulfills our condition. Therefore the general $C_\epsilon$ would be $$\Large C_\epsilon=\max\left\{\dfrac{1}{\left(1-(1-\epsilon)^{\frac{1}{p}}\right)^p},\dfrac{1+\epsilon}{((1+\epsilon)^{\frac{1}{p}}-1)^p}\right\}$$

Answer 2

Here is another solution based on Jensen's inequality.

We first prove that for any $\varepsilon>0$, there is $C_\varepsilon(p)>0$ such that $$\begin{align}|a+b|^p\leq (1+\varepsilon)|a|^p+C_\varepsilon(p)|b|^p\tag{0}\label{zero} \end{align}$$

• Suppose $p>1$. As the function $t\mapsto |t|^p$ is convex, for any $0<\lambda <1$ $$\begin{align} |a+b|^p\leq(|a|+|b|)^p&=\Big(\lambda\frac{|a|}{\lambda}+(1-\lambda)\frac{|b|}{1-\lambda}\Big)^p\\ &\leq \lambda\frac{|a|^p}{\lambda^p}+(1-\lambda)\frac{|b|^p}{(1-\lambda)^p}\\ &=\lambda^{1-p}|a|^p+(1-\lambda)^{1-p}|b|^p \end{align} $$ The choice $\lambda=(1+\varepsilon)^{-\tfrac{1}{p-1}}$ yields $$|a+b|^p\leq(1+\varepsilon)|a|^p +C_\varepsilon|b|^p$$ with $$C_\varepsilon=\frac{1+\varepsilon}{\Big((1+\varepsilon)^{\tfrac{1}{p-1}} -1\Big)^{p-1}}$$

• For $0<p\leq 1$, the inequality follows from the trivial inequality $(1+x^p)\leq 1+ x^p$ (for in this case, the map $t\mapsto (1+t)^p-t^p$, $t\geq0$, is nonincreasing). Then the statement of the OP holds with $C_\varepsilon=1$ for all $\varepsilon>0$, i.e. $$|a+b|^p\leq|a|^p+|b|^p\leq (1+\varepsilon)|a|^p+C_\varepsilon|b|^p$$

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To conclude the problem, applying inequality \eqref{zero} yields $$|a|^p=|(a+b)-b|^p\leq (1+\varepsilon)|a+b|^p+C_\varepsilon|b|^p$$ whence $$\frac{1}{1+\varepsilon}|a|^p\leq |a+b|^p+\frac{C_\varepsilon}{1+\varepsilon}|b|^p\leq |a+b|^p+C_\varepsilon|b|^p.$$ If $0<\varepsilon<1$, then $1-\varepsilon<\frac{1}{1+\varepsilon}$ and so, $$(1-\varepsilon)|a|^p-C_\varepsilon|b|^p\leq|a+b|^p$$ If $1\leq\varepsilon$, then it is clear that $$(1-\varepsilon)|a|^p-C_\varepsilon|b|^p\leq |a+b|^p$$ Putting things together, we obtain $$\big||a+b|^p-|a|^p\big|\leq \varepsilon|a|^p+C_\varepsilon|b|^p$$

Answer 3

I've taken the idea from my hint (now gone), but instead of a proof by contradicition, I give a direct proof. It has elements in common with the accepted answer, but it may be simpler.

Lemma: There exists a positive $C_\epsilon \ge 1$ such that

$$\tag 1\left||z+1|^p-1\right| \le \epsilon + C_\epsilon|z|^p$$

for all $z\in \mathbb C.$

Proof: Since the left side $\to 0$ as $z\to 0,$ there exists $\delta >0$ such that

$$\left||z+1|^p-1\right| < \epsilon\,\text{ for } |z|< \delta.$$

For $|z|\ge \delta,$ note that

$$\tag 2 \frac{\left||z+1|^p-1\right|}{|z|^p}$$

is nonnegative, continuous, and $\to 1$ as $|z|\to \infty.$ It follows that $(2)$ is bounded on $\{|z|\ge \delta\}.$ Let $C_\epsilon$ be the supremum of $(2)$ on this domain. Then $C_\epsilon\ge 1.$ We then have the left side of $(1)$ less than $\epsilon$ on $\{|z|<\delta\}$ and above by $C_\epsilon|z|^p$ on $\{|z|\ge\delta\}.$ This gives the lemma.

Now to our problem,, where we are given $a,b\in \mathbb C.$ The $C_\epsilon$ found in the lemma will be the $C_\epsilon$ that works in the problem.

If $a=0,$ the inequality reduces to $|b|^p\le C_\epsilon |b|^p,$ which holds since $C_\epsilon\ge 1.$ Otherwise we can divide by $|a|^p$ and the desired inequality becomes

$$\left||(b/a)+1|^p-1\right| \le \epsilon + C_\epsilon|b/a|^p.$$

This holds by the lemma and we are done.

Answer 4

If $a=0$, the inequality $$ \left||a+b|^p - |a|^p \right| \leq \epsilon |a|^p + C_{\epsilon} |b|^p\tag1 $$ is true with $C_\epsilon=1$. Otherwise, divide by $|a|^p$ and set $x=\frac ba$ to get $$ \left||1+x|^p-1\right|\le\epsilon+C_\epsilon|x|^p\tag2 $$ We will let $u=x+1$, so that $(2)$ becomes $$ \left|1-|u|^p\right|\le\epsilon+C_\epsilon|1-u|^p\tag3 $$

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Case: $\boldsymbol{|u|\le1}$

In this case, $(3)$ becomes $$ 1-|u|^p\le\epsilon+C_\epsilon|1-u|^p\tag4 $$ If $|u|\ge(1-\epsilon)^{1/p}$, then $(4)$ is trivially satisfied for any $C_\epsilon$; so suppose that $|u|\lt(1-\epsilon)^{1/p}$. Then the triangle inequality says $$ \begin{align} |1-u|^p &\ge(1-|u|)^p\tag5 \end{align} $$ Therefore, $$ \begin{align} \frac{1-|u|^p}{|1-u|^p} &\le\frac{1-|u|^p}{|1-|u||^p}\tag6 \end{align} $$ Thus, for $t=|u|\in\left[0,(1-\epsilon)^{1/p}\right]$ $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\frac{1-t^p}{(1-t)^p} &=\frac{p\left(1-t^{p-1}\right)}{(1-t)^{p+1}}\tag{7a}\\ &\le0&\text{if }p\le1\tag{7b}\\[6pt] &\ge0&\text{if }p\ge1\tag{7c} \end{align} $$ For $p\le1$, $\frac{1-t^p}{(1-t)^p}$ is decreasing, so its max is $1$ at $t=0$.
For $p\ge1$, $\frac{1-t^p}{(1-t)^p}$ is increasing, so its max is $\frac\epsilon{\left(1-(1-\epsilon)^{1/p}\right)^p}$ at $t=(1-\epsilon)^{1/p}$.

Therefore, we can use $$ C_\epsilon=\left\{\begin{array}{cl} 1&\text{if }p\le1\\ \frac\epsilon{\left(1-(1-\epsilon)^{1/p}\right)^p}&\text{if }p\gt1 \end{array}\right.\tag8 $$

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Case: $\boldsymbol{|u|\ge1}$

In this case, $(3)$ becomes $$ |u|^p-1\le\epsilon+C_\epsilon|u-1|^p\tag9 $$ If $|u|\le(1+\epsilon)^{1/p}$, then $(9)$ is trivially satisfied for any $C_\epsilon$; so suppose that $|u|\gt(1+\epsilon)^{1/p}$. Then the triangle inequality says $$ \begin{align} |u-1|^p &\ge(|u|-1)^p\tag{10} \end{align} $$ Therefore, $$ \begin{align} \frac{|u|^p-1}{|u-1|^p} &\le\frac{|u|^p-1}{||u|-1|^p}\tag{11} \end{align} $$ Thus, for $t=|u|\ge(1+\epsilon)^{1/p}$ $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\frac{t^p-1}{(t-1)^p} &=\frac{p\left(1-t^{p-1}\right)}{(t-1)^{p+1}}\tag{12a}\\ &\le0&\text{if }p\ge1\tag{12b}\\[6pt] &\ge0&\text{if }p\le1\tag{12c} \end{align} $$ For $p\le1$, $\frac{t^p-1}{(t-1)^p}$ is increasing, so its sup is $1$ as $t\to\infty$.
For $p\ge1$, $\frac{t^p-1}{(t-1)^p}$ is decreasing, so its max is $\frac\epsilon{\left((1+\epsilon)^{1/p}-1\right)^p}$ at $t=(1+\epsilon)^{1/p}$.

Therefore, we can use $$ \bbox[5px,border:2px solid #C0A000]{C_\epsilon=\left\{\begin{array}{cl} 1&\text{if }p\le1\\ \frac\epsilon{\left((1+\epsilon)^{1/p}-1\right)^p}&\text{if }p\gt1 \end{array}\right.}\tag{13} $$

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Bernoulli's Inequality says that for $p\gt1$, $$ (1+\epsilon)^{1/p}-1\le1-(1-\epsilon)^{1/p}\tag{14} $$ so for all $u$, we can use $(13)$ for $C_\epsilon$.