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Triangle inequality is used in one context or the other in analysis. To list a few $$ \|x+y\| \leq \|x\| + \|y\| $$

$$ d(x,y) \leq d(x,z) + d(z,y) $$

$$ \mu(A \cup B) \leq \mu(A) + \mu(B) $$

What is the motivation for imposing this? Intutively, it feels if we do not impose triangle inequality, the space will collapse to a trivial space in some sense. I am not able to express this formally. I would appreciate if someone could could throw some light on this and the motivation for triangle inequality?

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Well, distances between points in the plane and in space do satisfy the inequality... and the inequality is unmistakenly both useful and true in many contexts. – Mariano Suárez-Alvarez May 6 '11 at 17:30
If anything, I feel that without the triangle inequality the space would expand. – Zev Chonoles May 6 '11 at 17:35
@Mariano: Is that the sole reason? So what happens when we consider spaces where triangle inequality is not satisfied? – user17762 May 6 '11 at 17:36
@Rasmus: try your reversed triangle inequality with $x=y$. There are, however, other types of "reversed triangle inequality" that are nontrivial. – Robert Israel May 6 '11 at 18:18
I don't see much difference between 1. and 2. Subadditivity $\mu(A \cup B) \leq \mu(A) + \mu(B)$ has another motivation: the volume of a union can at most be the volume of the two parts. However, if you look at measurable sets modulo null sets (and stipulate $\mu$ finite), you get an honest distance $d_{\mu}(A,B) = \mu(A \triangle B)$ (symmetric difference on the right hand side). The associated metric space is quite useful in ergodic theory. Another nice feature of this space is that it is complete if and only if $\mu$ is complete. – t.b. May 6 '11 at 21:20
up vote 16 down vote accepted

Any reasonable notion of distance satisfies the triangle inequality, since if you can get from point $A$ to point $B$ using a route of length $d(A, B)$ and from point $B$ to point $C$ using a route of length $d(B, C)$, you can clearly get from point $A$ to point $C$ using a route of length $d(A, B) + d(B, C)$, and the optimal route (of length $d(A, C)$) therefore can't be any longer than this. (This argument is completely rigorous in an intrinsic metric, but there's no reason not to take it as motivation in general.)

Lawvere noticed that the above argument looks an awful lot like composition of morphisms in a category, and that led him to the categorical definition of metric spaces as certain kinds of enriched categories. From this perspective neither the symmetry nor the positive-definiteness axioms are particularly natural, but the triangle inequality is still very natural.

@Rasmus: letting $x = y$ in the reverse triangle inequality gives $0 \ge 2d(x, z)$, so the only "reverse metric space" is a point.

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I gather that Bregman divergence is a trendy thing to use in some geometric computational settings. It is supposedly a distance-like function that does not satisfy the triangle inequality. – yasmar May 6 '11 at 20:36

It seems I can't add comments so I decided to post this as an answer, I apologize if it seems trivial or redundant. My point of view is that most occurrences of this inequalities are just minimum principles (something you could summarize as "min $\le$ any"). In each case, the quantity (distance, norm, dimension, cardinal...) can be viewed as a minimum :

  • The distance between two points is the length of the shortest path between them

  • The dimension of a vector space smallest size for a system of generator

  • The cardinal of a set is the smallest $n$ for which your set can be embedded in $[\!|1,n|\!]$


In that setting, the triangle inequality more or less states that the right hand side correspond to a possibility (of path, generating set, embedding...) :

  • A path from $x$ to $z$ and then $z$ to $y$ is a path from $x$ to $y$.

  • The union of a generating sets gives a generating set for the sum.

  • If you can embed $A$ in $A'$ and $B$ in $B'$ then $A \cup B$ can be embedded in the disjoint union of $A'$ and $B'$


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In the context of normed spaces, subadditivity ensures local convexity, and that the Hahn-Banach theorem can be applied. For example, $L^p[0,1]$ with $0<p<1$ is not locally convex, and worse, it has no nonzero continuous linear functionals. If you try to define a "norm" by $\|f\|_p = \left(\int |f|^p\right)^{1/p}$, it is not subadditive, although it does satisfy $\|\lambda f\|_p=|\lambda|\|f\|_p$ if $\lambda$ is scalar. You can instead define a metric by $d(f,g)=\int |f-g|^p$, which gives $L^p$ the structure of an F-space. The triangle inequality is thus fixed, but the scalars no longer come out nice, and local convexity is lost regardless.

There is in fact a "reverse Minkowski inequality": if $f$ and $g$ in $L^p$ are nonnegative, $0<p<1$, then $\|f\|_p+\|g\|_p\leq \|f+g\|_p$.

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