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Let $a_1, a_2, \cdots , a_n$, $b_1, b_2, \cdots, b_n$ be real numbers, $p \in [1, +\infty)$, prove that

$$\sum_{1\leq i < j \leq n} |a_ib_j - a_jb_i|^p \leq c_p \sum_{i=1}^n |a_i|^p \sum_{i=1}^n |b_i|^p$$

where $c_p = \max(1, 2^{p-2})$.

Remark:

  1. When $p=1$, the proof is straightforward since $|a_ib_j-a_jb_i| \leq |a_i||b_j| + |a_j||b_i|$, and summing up all these inequalities is enough.

  2. When $p=2$, the inequality is a direct consequence of Lagrange's identity: $RHS - LHS = (\sum_{i=1}^n a_i b_i)^2 \geq 0$

  3. I have generated millions of sets of random $(a_i, b_i)$, for various values of $p$, and didn't find a counterexample.

Thanks!

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    $\begingroup$ Have you tried to generalize to complex numbers and mimic the proof of Riesz-Thorin interpolation theorem? (interpolating between $1$ and $2$ and then between $2$ and $\infty$) $\endgroup$ – fedja Apr 28 '18 at 1:49
  • $\begingroup$ @fedja : Thanks for the reference! Do you mean trying to prove the log convexity (in 1/p) of LHS/RHS? If so, I just verified numerically that it isn't. Also I am not sure about how to generalize to complex numbers. $\endgroup$ – Chen Dan Apr 28 '18 at 4:23
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    $\begingroup$ It may be worth mentioning that the main difficulty is in the sharp constant $c_p$. The same inequality with $c_p = 2^{p-1}$ follows from the elementary inequality on positive numbers $(x + y)^p \leq 2^{p-1}(x^p + y^p)$. $\endgroup$ – Willie Wong Apr 30 '18 at 13:34
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Note: the answer here stems from the comment of user fedja above.

The following Lemma is another version of the Riesz–Thorin interpolation theorem (See Lemma 8.5 in this book).

Lemma. Let $(X_i , \mathfrak M_i , μ_i )$, $i = 0,1, 2, . . . , n$ be measure spaces. Let $V_i$ represent the complex vector space of simple functions on $X_i$. Suppose that $$\Lambda : V_1\times V_2 \times \cdots \times V_n \to V_0$$ is a multilinear operator of types $p_0$ and $p_1$ where $p_0,p_1 \in [1,\infty]$, with constants $C_0$ and $C_1$, respectively. i.e., $$ \|\Lambda(f_1 , f_2 , . . . , f_n) \|_{p_i} \leq C_i \|f_1\|_{p_i} \|f_2 \|_{p_i} \cdots \|f_n\|_{p_i} \tag{1} $$ for $i=0,1$. Let $t \in [0, 1]$ and define $$\frac 1{p_t} := \frac{1-t}{p_0}+\frac t{p_1}$$ Then, $\Lambda$ is of type $p_t$ with constant $C_t :=C_0^{t-1}C_1^t$, that is, (1) holds true for $i=t$.

Now, Let $(X_i , \mathfrak M_i , μ_i )$ be a uniform measure on $[n] := \{1,\ldots,n\}$ for $i=1,2$, and be a uniform measure on $[n^2]$ for $i=0$. In this case, we have $V_1= V_2=\mathbb{C}^n$, and $V_0 = \mathbb{C}^{n^2}$. Define $\Lambda : V_1\times V_2 \to V_0$, by $[\Lambda(a,b)]_{i,j} := a_ib_j -a_jb_i$ for $a,b\in \mathbb{C}^n$ and $1\leq i,j\leq n$. Note that, in this setting, we have

$$ \|\Lambda(a,b)\|_p^p = \frac 2{n^2} \sum_{1\leq i<j\leq n} |a_i b_j - a_j a_i|^p\\ \|a\|_p^p = \frac 1n \sum_i |a_i|_p, \quad \|b\|_p^p = \frac 1n \sum_i |b_i|_p $$

Next, $\Lambda$ is of type $p_0=0$ and $p_1=1$, with constants $C_0=C_1=2$. Above Lemma implies that $\Lambda$ is of type $p$, for every $1<p<2$, with constant $2$.

Also $\Lambda$ is of type $p_0=2$,and $p_1=\infty$, with constant $C_0=2$ and $C_1=4$, respectively. Above Lemma implies that $\Lambda$ is of type $p$, for every $2<p<\infty$, with constant $2 \times 2^{(p-2)/p}$.

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  • $\begingroup$ Thanks, Mahdi! There is a generalization of the inequality, however I find it difficult to use your proof there - I asked a new question on MO: mathoverflow.net/questions/304076/… $\endgroup$ – Chen Dan Jul 2 '18 at 0:39

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