I have a small question that I think is very basic but I am unsure how to tackle since my background in computing inequalities is embarrassingly weak -

I would like to show that, for a real number $p \geq 1$ and complex numbers $\alpha, \beta$, I have \begin{equation} |\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p) \end{equation}

I thought it would be best to rewrite this as \begin{equation} \left|\frac{\alpha + \beta}{2}\right|^p \leq \frac{|\alpha|^p + |\beta|^p}{2} \end{equation}

but then I am unsure what to do next - is this a sensible start anyways ? Any help would be great !

(P.S. this is not a homework question - I am currently trying to brush up my knowledge of $L^p$ spaces, and this inequality came up as a statement. I thought it might be worthwhile to make sure I can fill in the gaps to improve my skills in computing inequalities.)

  • 1
    $\begingroup$ The following may be quite useful: $(a-b)^2\geq 0$ for real $a,b$. As well, the triangle inequality is worth considering. $\endgroup$
    – Alex R.
    Commented May 9, 2012 at 19:15
  • $\begingroup$ You could use convexity and Jensen's inequality in some way or another. $\endgroup$ Commented May 9, 2012 at 19:16
  • 1
    $\begingroup$ Check Minkowski's inequality . $\endgroup$
    – Theorem
    Commented May 9, 2012 at 19:27
  • $\begingroup$ thank you all for your comments! Jensen's as well as Minkowski's inequality are coming in the sequel of the book I am using ( "Analysis" by Lieb) so I guess I need to be able to derive this without using them .. $\endgroup$
    – harlekin
    Commented May 9, 2012 at 19:34
  • $\begingroup$ Let me know if you have any problems! $\endgroup$
    – Theorem
    Commented May 9, 2012 at 19:36

5 Answers 5

  • The map $x\mapsto x^p$ for $x\geq 0$ is convex, since its second derivative is $p(p-1)x^{p-2}\geq 0$.
  • We have $$\left|\frac{a+b}2\right|^p\leq \left(\frac{|a|+|b|}2\right)^p\leq \frac{|a|^p+|b|^p}2.$$
  • $\begingroup$ loved ur proof ! very short indeed. $\endgroup$
    – Theorem
    Commented May 9, 2012 at 19:39
  • 6
    $\begingroup$ It might be informative to mention Jensen's Inequality. $\endgroup$
    – robjohn
    Commented May 9, 2012 at 20:26
  • $\begingroup$ Can this inequality be extended to an infinite sum? I am interested in the case of a countable sum $\endgroup$
    – darkside
    Commented Feb 26 at 12:12

Without loss of generality, we can assume that that $\lvert \alpha \rvert \geq \lvert \beta \rvert$. Now essentially you want to prove that $$\displaystyle \left \lVert \frac{z + 1}{2} \right \Vert^p \leq \displaystyle \frac{\lVert z \rVert^p + 1}{2},$$ where $\displaystyle z = \frac{\alpha}{\beta}$ and $\lVert z \rVert \geq 1$.

Note that $$ \displaystyle \left \lVert \frac{z + 1}{2} \right \Vert \leq \frac{\lVert z \rVert + 1}{2}$$

So if we prove that $\displaystyle \left(\frac{1+t}{2} \right)^p \leq \frac{1+t^p}{2}$, where $t = \lVert z \rVert \geq 1$, we are done.

Now this is a one-variable calculus problem. Consider $f(t) = \displaystyle \frac{1+t^p}{2} - \left(\frac{1+t}{2} \right)^p$. We then get that $$\displaystyle f'(t) = \frac{p t^{p-1}}{2} - \frac{p (1+t)^{p-1}}{2^p}.$$ Hence, $$f'(t) = \frac{p}{2} \left( t^{p-1} - \left(\frac{1+t}{2} \right)^{p-1}\right) \geq 0$$ for $t \geq 1$. Hence, $f(t)$ is increasing for all $t \geq 1$. And $f(1) = 0$. Hence, we have that $$f(t) \geq f(1) = 0.$$ Hence, we get that $$\displaystyle \left(\frac{1+t}{2} \right)^p \leq \frac{1+t^p}{2},$$ where $t = \lVert z \rVert \geq 1$


In fact, one advantage of the argument in john w.'s answer is that it can be extended further to prove $$(a+b)^n\leq p^na^n+q^nb^n,$$ where $\frac{1}{p}+\frac{1}{q}=1$. In particular, one can choose the coefficient of $a^n$ very close to $1$ by paying the price of a larger coefficient of $b^n$.

The proof is achieved by arguing either $(a+b)\leq pa$ or $(a+b)\leq qb$ holds. This is true since otherwise one would get a contradiction that $\left(\frac{1}{p}+\frac{1}{q}\right)(a+b)>a+b$.

  • $\begingroup$ I suppose your "contradiction" $\left(\frac{1}{p}+\frac{1}{q}\right)(a+b)\color{red}{>}a+b$ is not valid! $\endgroup$
    – MathArt
    Commented Mar 11, 2022 at 8:34

If the $p-1$ being $p$ is not a big deal then I think the following works. $|\alpha+\beta|\leq |\alpha|+|\beta|\leq 2\max\{|\alpha|,|\beta|\}$. If the max is $|\alpha|$, then $|\alpha+\beta|^p\leq 2^p|\alpha|^p\leq 2^p(|\alpha|^p+|\beta|^p)$. Similarly for $\beta$. In either case you get the inequality you want.

  • $\begingroup$ It is actually a big deal, the inequality making use the convexity normally only overestimates the differences half as the inequality choosing the biggest one. $\endgroup$
    – Shuhao Cao
    Commented May 13, 2012 at 2:01

To generalize this $L_p$ (p>1) norm for $\alpha$ and $\beta$, we can introduce weights such that $q+r=1$, so we have \begin{align} q\alpha+r\beta=(q\alpha^p)^{1/p}q^{1-1/p}+(r\beta^p)^{1/p}r^{1-1/p}\underset{Holder}{\le}\left(q\alpha^p+r\beta^p\right)^{1/p}(q+r)^{1-1/p}=\left(q\alpha^p+r\beta^p\right)^{1/p} \end{align} therefore, \begin{align} (q\alpha+r\beta)^p\le q\alpha^p+r\beta^p. \end{align} In case of $q=r=\frac12$, the originally asked is then obtained. \begin{align} \left|\frac{\alpha + \beta}{2}\right|^p \leq \frac{|\alpha|^p + |\beta|^p}{2} \end{align} Note the sign "$\le$" becomes "$\ge$" for the case $0<p<1$.


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