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I've been playing around with an extension of a dot product to three vectors, as set forth in this question. Basically, if you have three vectors A, B, and C, then you could compute the following

$$TD(A,B,C) = \sum_{i=1}^N a_i b_i c_i$$

where TD means "triple dot." I realize this isn't an accepted notation but it is useful for this question.

I'm curious if anyone has shown that

$$TD(A,B,C)\leqslant \lVert A \rVert \cdot\lVert B \rVert\cdot \lVert C \rVert.$$

I suppose it might involve an extension of the Rearrangement inequality to three vectors.

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  • $\begingroup$ @C. Falcon, I'd appreciate it if you could undelete your previous answer. Even if there was a problem with it, it might inspire others to fix it. $\endgroup$ – gaefan Dec 30 '15 at 22:44
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Upon Jeff's request here the solution I provided. The validity of the proof as it is right now requires that: $$\sum_{1\leqslant 1<i<j\leqslant n}b_ic_j\geqslant 0.$$


Using Lagrange's identity, for $(a_i)$ and $(b_ic_i)$, one has: $$\textrm{TD}(A,B,C)^2=\|A\|^2\cdot\sum_{i=1}^n(b_ic_i)^2-\underbrace{\sum_{1\leqslant i,j\leqslant n}^n(a_ib_jc_j-a_jc_ib_i)^2}_{\geqslant 0}.$$ Hence, one derives: $$\textrm{TD}(A,B,C)^2\leqslant\|A\|^2\cdot\sum_{i=1}^n(b_ic_i)^2.\tag{1}$$ Besides, one has: $$\langle B,C\rangle^2=\sum_{i=1}^n(b_ic_i)^2+\underbrace{2\sum_{1\leqslant i<j\leqslant n}b_ic_j}_{\geqslant 0}.$$ Therefore, one derives: $$\sum_{i=1}^n(b_ic_i)^2\leqslant\langle B,C\rangle^2.\tag{2}$$ Using $(1)$ and $(2)$, one has: $$|\textrm{TD}(A,B,C)|\leqslant \|A\|\cdot|\langle B,C\rangle|.\tag{3}$$ Finally, using Cauchy-Schwarz inequality, one gets: $$|\textrm{TD}(A,B,C)|\leqslant\|A\|\cdot\|B\|\cdot\|C\|.$$

Remark. Inequality $(3)$ is stronger than the one you are interested in.

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  • $\begingroup$ Cool! How does the first inequality follow from Lagrange's identity? The wikipedia entry states it as something involving subtraction. $\endgroup$ – Eli Rose Dec 30 '15 at 19:52
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A short proof can be given using the Frobenius norm: As already pointed out in the question you linked to, we have $$ \mathrm{TD}(a,b,c) = a^T \cdot \mathrm{diag}(b_1, \dotsc, b_n) \cdot c $$ where $\mathrm{diag}(b_1, \dotsc, b_n)$ denotes the diagonal matrix with diagonal entries $b_1, \dotsc, b_n$. Using the submultiplicativity of the Frobenius norm we find that \begin{align*} \mathrm{TD}(a,b,c) &\leq |\mathrm{TD}(a,b,c)| = \|\mathrm{TD}(a,b,c)\|_F = \|a^T \cdot \mathrm{diag}(b_1, \dotsc, b_n) \cdot c\|_F \\ &\leq \|a^T\|_F \cdot \|\mathrm{diag}(b_1, \dotsc, b_n)\|_F \cdot \|c\|_F = \|a\| \cdot \|b\| \cdot \|c\|, \end{align*} where $\|\cdot\|_F$ denotes the Frobenius norm.

Another proof can be given by using the Cauchy-Schwarz inequality: We can assume w.l.o.g. that $a_i, b_i, c_i \geq 0$ for every $1 \leq i \leq n$. Because $b_i \geq 0$ for all $1 \leq i \leq n$ we find that the bilinear form $\langle \cdot, \cdot \rangle_b$ defined via $$ \langle x,y \rangle_b = x^T \cdot \mathrm{diag}(b_1, \dotsc, b_n) \cdot y = \mathrm{TD}(x,b,y) $$ is symmetric and positive semidefinite with $$ \|x\|_b = \sqrt{\langle x,x \rangle_b} = \sqrt{ \sum_{i=1}^n b_i x_i^2 }. $$ Notice that for all $1 \leq i,k \leq n$ we have $b_i b_k \leq \sum_{j=1}^n b_j^2$: If $i = k$ this is clear and if $i \neq k$ then $$ b_i b_k \leq 2 b_i b_k \leq b_i^2 + b_k^2 \leq \sum_{j=1}^n b_j^2. $$ So from the Cauchy-Schwarz inequality it follows that \begin{align*} \mathrm{TD}(a,b,c) &= \langle a, c \rangle_b \leq \|a\|_b \cdot \|c\|_b = \sqrt{ \sum_{i=1}^n b_i a_i^2} \cdot \sqrt{\sum_{k=1}^n b_k c_k^2 } \\ &= \sqrt{ \sum_{i,k=1}^n a_i^2 b_i b_k c_k^2} \leq \sqrt{ \sum_{i,j,k=1}^n a_i^2 b_j^2 c_k^2 } \\ &= \sqrt{\sum_{i=1}^n a_i^2} \cdot \sqrt{\sum_{j=1}^n b_j^2} \cdot \sqrt{\sum_{k=1}^n c_k^2} = \|a\| \cdot \|b\| \cdot \|c\|. \end{align*}

It is worth noticing that using the approach via the Frobenius norm we can also directly generalize our results to arbitrary $x^1, \dotsc, x^m \in \mathbb{C}^n$, in the sense that \begin{align*} \left|\sum_{i=1}^n x^1_i \dotsm x^m_i\right| &= \left\| (x^1)^T \cdot \mathrm{diag}(x^2_1, \dotsc, x^2_n) \dotsm \mathrm{diag}(x^{m-1}_1, \dotsc, x^{m-1}_n) \cdot x^m \right\|_F \\ &\leq \|(x^1)^T\|_F \cdot \|\mathrm{diag}(x^2_1, \dotsc, x^2_n)\|_F \dotsm \|\mathrm{diag}(x^{m-1}_1, \dotsc, x^{m-1}_n)\|_F \cdot \|x^m\|_F \\ &= \|x^1\| \dotsm \|x^m\|. \end{align*}

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  • $\begingroup$ Thank you for the detailed answer! Why do you say "using that a_i,b_i,c_i≥0". It seems that these can be negative. $\endgroup$ – gaefan Dec 30 '15 at 22:42
  • $\begingroup$ My bad. I did some reshuffeling of the paragraphs before posting the answer and forgot to remove this. $\endgroup$ – Jendrik Stelzner Dec 30 '15 at 22:44

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