We know about $L_p$ spaces that average of the $p$-th powers of a vector is a norm:

$$ |x|_p = \sqrt[p]{ \frac{x_1^p + \dots + x_n^p}{p}} $$

This means for example, this norm satisfies the triangle inequality (in this case, it's the Minkowski inequality):

$$ |x+y|_p = \sqrt[p]{ \frac{(x_1+y_1)^p + \dots + (x_n+y_n)^p}{p}} \leq \sqrt[p]{ \frac{x_1^p + \dots + x_n^p}{p}} + \sqrt[p]{ \frac{y_1^p + \dots + y_n^p}{p}} = |x|_p + |y|_p $$

I'd like to know if we could use other symmetric polynomials as norms. Let's define a function:

$$ f(x) = \sum_{1 \leq i < j < k \leq n} |x_i x_j x_k|^3 $$

Does this satisfy triangle inequality? E.g. is $f(x+y) < f(x) + f(y)$, where $x_i, y_i > 0$ ? And if it is a norm we typically have constants such that:

$$ A ||\cdot ||_1 < ||\cdot||_2 < B ||\cdot||_1 $$

and I'd like to know possible constants $A$ and $B$, not even optimal.

Originally my question was if $|abc|+|bcd|+|abd|+|acd|$ is a norm on vectors $(a,b,c,d) \in \mathbb{R}^4$ and to compare it to the $L^3$ norm.

Please put absolute value signs as needed. At least, what I've written s hould be true for $p$ even.

  • $\begingroup$ Did you mean $(x_n + y_n)^p$ in your version of Minkowski instead of $(x_n + n_1)^p$? Also, your expression for the $p$-norm isn't correct. $\endgroup$ – B. Mehta Jul 30 '18 at 20:28
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    $\begingroup$ It's not the $p$-norm (it is up to a factor $p^{1/p}$), but it is a norm (for $p\geq 1$). $\endgroup$ – Clement C. Jul 30 '18 at 20:31
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    $\begingroup$ your expressions with $x_ix_jx_k$ $i<j<k$ are 1) not $1$-homogeneous 2) not convex 3) are 0 for the canonical basis so they satisfy precisely 0 conditions out of 3 for being a norm. $\endgroup$ – Surb Jul 30 '18 at 20:42
  • $\begingroup$ What is your question? The question in the title is different from the (rather unclear) question in the body of your post. One very simple criterion for a norm is that it must restrict to a norm on any subspace. Your proposed norms on $\Bbb{R}^4$ do not restrict to a norm on the $x_1$-axis. $\endgroup$ – Rob Arthan Jul 30 '18 at 20:48

Let $p\geq 1$ and $$ f(x) = \sum_{1 \leq i < j < k \leq n} |x_i x_j x_k|^p $$ Then $f(1,0,\ldots,0)=0$ so it can not be a norm, let us fix that by redefining $f$ as $$ \tilde f(x) = \sum_{1 \leq i \leq j \leq k \leq n} |x_i x_j x_k|^p $$ Now, $\tilde f(\alpha x)= \alpha^{3p} f(x)$ for $\alpha>0$ so it is still not norm, let us fix this by redefining $\tilde f$ as $$ \hat f(x) = \Big(\sum_{1 \leq i \leq j \leq k \leq n} |x_i x_j x_k|^p\Big)^{\frac{1}{3p}}.$$ Now, by computing the Hessian of $\hat f$, you will see that it is not always positive semi-definite. Hence $\hat f$ is not convex and therefore not a norm (and I don't think this can be fixed).

Conclusion: $f$ is not a norm and not even close from being so.


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