📂Lebesgue Spaces

Proof of Clarkson's Inequalities

Theorem¹

Let’s say that $\Omega \subset \mathbb{R}^{n}$ is an open set.

Let’s say that $u,v\in {L}^{p}(\Omega)$. Also, let’s say it satisfies $\frac{1}{p}+\frac{1}{p^{\prime}}=1$. If $2 \le p \lt \infty$, then the following two inequalities hold.

$$ \begin{equation} \left\| \frac{u+v}{2}\right\|_{p}^{p}+ \left\| \frac{u-v}{2} \right\|_{p}^{p} \le \frac{1}{2}\left\| u \right\|_{p}^{p} + \frac{1}{2}\left\| v \right\|_{p}^{p} \end{equation} $$

$$ \begin{equation} \left\| \frac{u+v}{2}\right\|_{p}^{p^{\prime}}+ \left\| \frac{u-v}{2} \right\|_{p}^{p^{\prime}} \ge \left( \frac{1}{2}\left\| u \right\|_{p}^{p} + \frac{1}{2}\left\| v \right\|_{p}^{p}\right)^{p^{\prime}-1} \end{equation} $$

If $1 \lt p \le 2$, then the following two inequalities hold.

$$ \begin{equation} \left\| \frac{u+v}{2}\right\|_{p}^{p}+ \left\| \frac{u-v}{2} \right\|_{p}^{p} \ge \frac{1}{2}\left\| u \right\|_{p}^{p} + \frac{1}{2}\left\| v \right\|_{p}^{p} \end{equation} $$

$$ \begin{equation} \left\| \frac{u+v}{2}\right\|_{p}^{p^{\prime}}+ \left\| \frac{u-v}{2} \right\|_{p}^{p^{\prime}} \le \left( \frac{1}{2}\left\| u \right\|_{p}^{p} + \frac{1}{2}\left\| v \right\|_{p}^{p}\right)^{p^{\prime}-1} \end{equation} $$

Explanation

This is called Clarkson’s inequalities.

$(1)$ and $(3)$, $(2)$ and $(4)$ have opposite sign directions in the same equation.

As with many inequalities, rather than having significant meaning by themselves, they are handy in proving other important theorems. As there are many auxiliary theorems and formulas needed for the proof, we will organize them in advance. Since $\frac{1}{p}+\frac{1}{p^{\prime}}=1$, we can obtain the following formulas when organized well.

$$ \begin{array}{c} \displaystyle (a)\ p^{\prime}=p(p^{\prime}-1),\quad (b)\ p=p^{\prime}(p-1),\quad (c)\ \frac{p}{p^{\prime}}=p-1 \\ \displaystyle (d)\ \frac{p^{\prime}}{p}=p^{\prime}-1,\quad (e)\frac{1}{p-1}=p^{\prime}-1\end{array} \quad \cdots (5) $$

Also, using the definition of a norm, it is easy to see that the following formula holds for $u\in {L}^{p}$.

$$ \| u \|_{p}^{p^{\prime}}=|\ |u|^{p^{\prime}} |_{p-1}\quad \cdots (6) $$

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Auxiliary Theorem

Let’s say $z,w \in \mathbb{C}$ and $\frac{1}{p}+\frac{1}{p^{\prime}}=1$. If $1<p \le 2$, then the following inequality holds.

$$ \left| \frac{z+w}{2} \right|^{p^{\prime}} + \left| \frac{z-w}{2} \right|^{p^{\prime}} \le \left( \frac{1}{2}|z|^{p} + \frac{1}{2}|w|^{p} \right)^{\frac{1}{p-1}}\quad \cdots (7) $$

If $2 \le p <\infty$, then the following inequality holds.

$$ \left| \frac{z+w}{2} \right|^{p} + \left| \frac{z-w}{2} \right|^{p} \le \frac{1}{2}|z|^{p} + \frac{1}{2}|w|^{p} \quad \cdots (8) $$

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Minkowski’s inequality

Let’s say $1 \le p<\infty$. If $u,v\in {L}^{ p}$, then the following inequality holds.

$$ | u+v|_{p} \le \| u \|_{p}+\left\| v \right\|_{p} $$

Reverse Minkowski’s inequality

Let’s say $0<p \le 1$. If $u,v\in {L}^{ p}$, then the following inequality holds.

$$ | \ |u|+|v| \ |_{p} \ge \| u \|_{p}+\left\| v \right\|_{p} $$

Proof

(1)

Assuming $2 \le p <\infty$. Inserting $z=u(x)$, $w=v(x)$ into $(8)$ and taking the integral, we directly get $(1)$.

$$ \int_{\Omega} \left( \left| \frac{u+v}{2} \right|^{p}+ \left| \frac{u-v}{2} \right|^{p} \right) dx \le \int_{\Omega} \left( \frac{1}{2}|u|^{p} +\frac{1}{2}|v|^{p} \right) dx $$

$$ \implies \int_{\Omega} \left| \frac{u+v}{2} \right|^{p}dx+ \int_{\Omega} \left| \frac{u-v}{2} \right|^{p} dx \le \frac{1}{2}\int_{\Omega} |u|^{p}dx +\frac{1}{2}\int_{\Omega} |v|^{p} dx $$

$$ \implies \left\| \frac{u+v}{2} \right\|_{p}^{p} + \left\| \frac{u-v}{2} \right\|_{p}^{p} \le \frac{1}{2}\| u \|_{p}^{p} + \frac{1}{2} \left\| v \right\|_{p}^{p} $$

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(2)

Assuming $2 \le p <\infty$. Then $1<p^{\prime}\le 2$ and $1\le p$. Using the results organized above for calculation

$$ \begin{align*} \left\| \frac{u+v}{2} \right\|_{p}^{p^{\prime}}+\left\| \frac{u+v}{2} \right\|_{p}^{p^{\prime}} =&\ \left\| \ \left| \frac{u+v}{2}\right|^{p^{\prime}} \right\|_{p-1}+\left\| \ \left| \frac{u-v}{2}\right|^{p^{\prime}} \right\|_{p-1} \\ \ge& \left\| \ \left|\frac{u+v}{2}\right|^{p^{\prime}} + \left| \frac{u-v}{2}\right|^{p^{\prime}} \right\|_{p-1} \\ =&\ \left( \int_{\Omega} \left| \ \left|\frac{u+v}{2}\right|^{p^{\prime}} + \left| \frac{u-v}{2}\right|^{p^{\prime}} \right|^{p-1} dx \right)^{\frac{1}{p-1}} \\ \ge& \left( \int_{\Omega} \left( \frac{1}{2} \left| u \right|^{p} + \frac{1}{2}\left| v \right|^{p} \right) dx \right)^{\frac{1}{p-1}} \\ =&\ \left( \frac{1}{2}\int_{\Omega} \left| u \right|^{p} dx + \frac{1}{2} \int_{\Omega} \left| v \right|^{p} dx \right)^{\frac{1}{p-1}} \\ =&\ \left( \frac{1}{2} \left\| u \right\|_{p}^{p} + \frac{1}{2} \left\| v \right\|_{p}^{p} \right)^{p^{\prime}-1} \end{align*} $$

The first line holds by $(6)$. The second line holds by the Minkowski inequality because of $1\le p-1$. The third line is the direct application of the definition of $p-1$ norm. More calculation is needed to prove the fourth line holds.

• Additional calculation Since $1<p^{\prime}\le 2$, inserting $p=p^{\prime}$, $z=u+v$, $w=u-v$ into $(7)$

  $$ \left| u \right|^{p} + \left| v \right|^{p} \le \left( \frac{1}{2}|u+v|^{p^{\prime}} + \frac{1}{2}|u-v|^{p^{\prime}} \right)^{\frac{1}{p^{\prime}-1}} $$

  Multiplying both sides by $\frac{1}{2}$

  $$ \begin{align*} \frac{1}{2}\left| u \right|^{p} + \frac{1}{2}\left| v \right|^{p} \le& \frac{1}{2}\left( \frac{1}{2}|u+v|^{p^{\prime}} + \frac{1}{2}|u-v|^{p^{\prime}} \right)^{\frac{1}{p^{\prime}-1}} \\ =&\ \left(\frac{1}{2}\right)^{{p^{\prime}-1} \frac{1}{p^{\prime}-1}} \left( \frac{1}{2}|u+v|^{p^{\prime}} + \frac{1}{2}|u-v|^{p^{\prime}} \right)^{p-1} \\ =&\ \left(\frac{1}{2}\right)^{(p^{\prime}-1) (p-1)} \left( \frac{1}{2}|u+v|^{p^{\prime}} + \frac{1}{2}|u-v|^{p^{\prime}} \right)^{p-1} \\ =&\ \left(\frac{1}{2^{p^{\prime}}}|u+v|^{p^{\prime}} + \frac{1}{2^{p^{\prime}}}|u-v|^{p^{\prime}} \right)^{p-1} \\ =&\ \left(\frac{u+v}{2}|^{p^{\prime}} + |\frac{u-v}{2}|^{p^{\prime}} \right)^{p- 1} \end{align*} \quad \cdots (9) $$

  The third line holds by $(5)(e)$. The fourth line holds by $(9)$. By organizing and using the definition of the norm and applying $(5)(e)$, the last line holds.

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(4)

Let’s say $1<p\le 2$.

$$ \begin{align*} \left\| \frac{u+v}{2} \right\|_{p}^{p^{\prime}}+\left\| \frac{u+v}{2} \right\|_{p}^{p^{\prime}} =&\ \left\| \ \left| \frac{u+v}{2}\right|^{p^{\prime}} \right\|_{p-1}+\left\| \ \left| \frac{u-v}{2}\right|^{p^{\prime}} \right\|_{p-1} \\ \le& \left\| \ \left|\frac{u+v}{2}\right|^{p^{\prime}} + \left| \frac{u-v}{2}\right|^{p^{\prime}} \right\|_{p-1} \\ =&\ \left( \int_{\Omega} \left|\ \left|\frac{u+v}{2}\right|^{p^{\prime}} + \left| \frac{u-v}{2}\right|^{p^{\prime}} \right|^{p-1} dx \right)^{\frac{1}{p-1}} \\ \le& \left( \int_{\Omega} \left| \left( \frac{1}{2} \left| u \right|^{p} + \frac{1}{2}\left| v \right|^{p}\right)^{\frac{1}{p-1}} \right|^{p-1} dx \right)^{\frac{1}{p-1}} \\ =&\ \left( \frac{1}{2}\int_{\Omega} \left| u \right|^{p} dx + \frac{1}{2} \int_{\Omega} \left| v \right|^{p} dx \right)^{p^{\prime}-1} \\ =&\ \left( \frac{1}{2} \left\| u \right\|_{p}^{p} + \frac{1}{2} \left\| v \right\|_{p}^{p} \right)^{p^{\prime}-1} \end{align*} $$

The first line holds by $(6)$. The second line holds by the reverse Minkowski inequality because of $0<p-1\le 1$. The fourth line holds by $(7)$. Using the definition of the norm and $(5)(e)$, the last line holds.

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