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Let $1<p<2$. I'm trying to prove the inequality

$$ |a+b|^q+|a-b|^q\leq 2\big( |a|^p + |b|^p \big)^{q-1} $$ where $\frac{1}{p}+\frac{1}{q}=1$.

Following this paper, I am able to prove the inequality for real $a,b$.

I'm missing the step to complex.

I have tried to set $x=|a+b|^2$ and $y=|a-b|^2$ (and other combinations) and then use the real inequality for $x,y$ with no success.

This is not exactly a duplicate of On the second Clarkson's inequality because that one asked from a "from scratch" proof while I'm only asking for one step.

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  • $\begingroup$ I found this answer which makes the work. One can found a proof of the required Hölder inequalities here (in French). If you are interested, I wrote a fully detailed proof in giulietta in the french part (search for "Clarkson"). $\endgroup$ May 7, 2019 at 20:10
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    $\begingroup$ I checked your document above (in French) and I think that the use of Holder's inequality in the proof of Clarkson's inequality (Lemma 27.136)is not done correctly there. The assumptions are $1<p<2$, thus the convex conjugate $q$ of $p$ is larger than $2$. Then you argue that $$\|(x+y, x-y\|^q_q\leq \Big(2^{1/q-1/2}\|x+y,x-y\|_2\big)^q$$ This is not quite right as $2<q$ (Holder's inequality goes in the opposite way: $0<r<s$ then $\|v\|_r\leq n^{1/r-1/2}\|v\|_s$ $\endgroup$
    – Mittens
    May 23, 2022 at 15:09
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    $\begingroup$ The inequality can be reduced to the real case by optimizing with respect to the angle between $a $ and $b$ , i.e. the argument of $\bar{a} b.$ The maximum of the left hand side is attained when the angle is $0$ or $\pi$ which means that one of the numbers is a real multiple of the other. $\endgroup$ May 24, 2022 at 5:54
  • $\begingroup$ @RyszardSzwarc: that is indeed the case, and that is done for example in Hewitt and Stromberg. The issue that I want to raise with to the OP is that his proof in the real case in his notes (see link before) is not correct. The real case appears also in Hewitt-Stomber and in Makarov, et. al, Selected problems in analysis, using Calculus methods, which are rather messy and tricky in this case. I just wanted to present an alternative proof which uses Riesz interpolation (which is usually covered in courses of graduate analysis) which is rather slick. $\endgroup$
    – Mittens
    May 24, 2022 at 16:12
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    $\begingroup$ @OliverDíaz Best, RS $\endgroup$ May 24, 2022 at 19:10

2 Answers 2

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Let's assume for simplicity that $a\neq 0$ and $b\neq 0.$ Then $$\displaylines{|a+b|^2=|a|^2+|b^2|+2\Re (a\bar b)= |a|^2+|b^2|+2|a|\,|b|\cos t\\ |a-b|^2=|a|^2+|b^2|-2\Re (a\bar b)= |a|^2+|b^2|-2|a|\,|b|\cos t }$$ Consider the function $f(t)$ on $[-\pi,\pi]$ $$\displaylines{f(t) =|a+b|^p+|a-b|^q\\ =\left (|a|^2+|b^2|+2|a|\,|b|\cos t\right )^{q/2}+\left (|a|^2+|b^2|-2|a|\,|b|\cos t\right)^{q/2}}$$ The function $f(t)$ is even, so we may restrict to $[0,\pi].$ We have $$\displaylines{f'(t)=-q\,|a|\,|b|\,\sin t\,\left [\left (|a|^2+|b^2|+2|a|\,|b|\cos t\right )^{q/2-1}\\ - \left (|a|^2+|b^2|-2|a|\,|b|\cos t\right )^{q/2-1}\right ]}$$ Moreover $f'(t)\le 0$ for $0\le t\le \pi/2$ and $f'(t)\ge 0$ for $\pi/2\le t\le \pi$ (as $q>2$). So the maximum is attained at $0$ or $\pi.$ Next $$f(0)=f(\pi)=(|a|+|b|)^q+\big |\, |a|-|b|\,\big |^q,\quad |a|\ge |b|$$ Summarizing, we have obtained $$|a+b|^q+|a-b|^q\le (|a|+|b|)^q+\big |\, |a|-|b|\,\big |^q $$ thus the problem has been reduced to real numbers.

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Since the inequality in question is used primarily to derive bounds in $L_p$, namely Clarkson's inequality for $1<p<2$: $$\Big\|\frac{f+g}{2}\Big\|^{p'}_p+\Big\|\frac{f-g}{2}\Big\|^{p'}_p\leq \Big(\frac12\big(\|f\|^p_p+\|g\|^p_p\Big)^{1/(p-1)},$$ I will present a short proof of the inequality in the OP for $\mathbb{C}$ based on interpolation methods, namely the Riesz-Thorin interpolation theorem which I state now:

Theorem: Let $(X,\mathscr{F},\mu)$ and $(Y,\mathscr{G},\nu)$ be $\sigma$-finite measures spaces and $1\leq p_0,p_1\leq\infty$. Suppose $T:L_{p_0}(\mu)+L_{p_1}(\mu)\rightarrow L_{q_0}(\nu)+ L_{q_1}(\nu)$ is a linear operator such that \begin{align} \|Tf\|_{q_0}&\leq M_0\|f\|_{p_0},\qquad f\in L_{p_0}(\mu)\\ \|Tg\|_{q_1}&\leq M_1\|g\|_{q_1},\qquad g\in L_{p_1}(\mu) \end{align} For any $0<t<1$ define $$\frac{1}{p_t}=\frac{1-t}{p_0}+\frac{t}{p_1},\qquad \frac{1}{q_t}=\frac{1-t}{q_0}+\frac{t}{q_1}$$ Then, $T:L_{p_t}(\mu)\rightarrow L_{q_t}(\nu)$ is a bounded linear operator and $$\|T\,h\|_{p_t}\leq M^{1-t}_0M^t_1\|h\|_{q_t}$$

Appealing to this theorem may appear as an overkill, however due to frat that the inequality of the OP is used primarily in the study of $L_p$ spaces, and that the interpolation theorem above is part of the curricula in integration theory, we might as well show a nice application of it. Also, elementary proof of the OP's inequality (using differentiation, or Calculus methods) are actually long and tricky, even in the real case (see for example, Hewitt, E. and Stronberg, K., Real and Abstract Analysis, Springer Verlag, GTM Vol 25, 1965, pp. 225-227). Finally, versions of the Riesz-Thorin's theorem in the context of $\ell_p$ also appear in the classic book Hardy, G. et. al., Inequalities, Cambridge University, 1934, pp. 217-219.


Proof of the OP's inequality: Consider the operator $T$ on $\mathbb{C}^2$ defined as $$T\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}a+b\\ a-b\end{pmatrix} $$ Since all $p$-norms on Eucliean space are equivalent, we may ask ourselves what is the operator norm $\|T\|_{p,q}$ when we consider $T:(\mathbb{C},\|\,\|_p)\rightarrow(\mathbb{C},\|\;\|_q)$. Notice that for $p=2=q$ $$|a+b|^2+|a-b|^2=2(|a|+|b|^2)$$ ans so $\|T\|_{2,2}=\sqrt{2}$. On the other hand, for $p=1$ and $q=\infty$ we have $\max\{|a+b|,|a-b|\}\leq|a|+|b|$ with equality when for example $a=b=1$. Thus $\|T\|_{1,\infty}=1$. Thus, for $p_t,q_t$ with $$\frac{1}{p_t}=\frac{1-t}{2}+t,\qquad \frac{1}{q_t}=\frac{1-t}{2}$$ $$\|T\|_{p_t,q_t}\leq 2^{(1-t)/2}$$ That is \begin{align} \Big(|a+b|^q+|a-b|^q\Big)^{1/q}\leq 2^{(1-t)/2}\big(|a|^p+|b|^p\big)^{1/p}\end{align} Notice that $\frac{1}{2}<\frac{1}{p}<1$ and that $\frac1p+\frac1q=1$. Substituting $q=\frac{2}{1-t}$ yields the desired inequality \begin{align} |a+b|^q+|a-b|^q\leq 2 \big(|a|^p+|b|^p\big)^{q/p}= 2 \big(|a|^p+|b|^p\big)^{q-1}\tag{1}\label{one}\end{align}


Comment: Clarkson's inequality $$ |a+b|^q+|a-b|^q\leq 2^{q-1}(|a|^q+|b|^q),\qquad q\geq2$$ can be obtained by interpolating between $\|T\|_{2,2}$ and $\|T\|_{\infty,\infty}$; another approach is to use \eqref{one} and Hólder's inequality: Since $\frac1p+\frac1q=1$, $1<p<q$ and so, $$ |a|^p+|b|^p=2\Big(\frac12 |a|^p\cdot1+\frac12 |b|^p\cdot1\Big)\leq 2\Big(\frac{1}{2}|a|^q+\frac{1}{2}|b|^q\Big)^{p/q}=2^{1-\tfrac{p}{q}}(|a|^q+|b|^q)^{p/q} $$ Putting things together, we obtain $$|a+b|^q+|a-b|^q\leq 2\big(|a|^p+|b|^p\big)^{q-1}\leq 2^{q/p}(|a|^q+|b|^q)=2^{q-1}(|a|^q+|b|^q) $$


The idea of using interpolation to derive a simple proof of Clarkson's inequalities for $\mathbb{C}$ appears in the paper Boas, R. P. Jr,, Some Uniformly Convex Spaces, Bull. Amer. Math. Soc. Vol.46 • No. 4 • April 1940., pp. 304-311

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