What purpose does Triangle inequality serve while defining norms?
Is it important to render it as a distance metric?
What would happen if we let the condition loose?
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7$\begingroup$ Without the triangle inequality, good luck proving things like if $x_n \to x$ and $y_n \to y$, then $x_n + y_n \to x + y$. $\endgroup$– D_SJul 15, 2017 at 1:23
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1$\begingroup$ Another way of informally stating the triangle inequality is "the shortest distance between two points is a straight line". As D_S points out, the concepts of limits and neighborhoods would be warped without it. $\endgroup$– fleabloodJul 15, 2017 at 1:25
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1$\begingroup$ @fleablood: But a line makes sense in Euclidean space, but, how do you interpret the inequality with lines in more general spaces? $\endgroup$– garyJul 15, 2017 at 1:34
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1$\begingroup$ @gary The "intuitive" structure of Euclidean space motivates how we choose to define other spaces. $\endgroup$– MathematicsStudent1122Jul 15, 2017 at 1:37
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1$\begingroup$ @fleablood: I somewhat agree, but there are some pretty wild spaces out there where lines do not make sense, i.e., space is not convex. I got a cold shower in a class on Metric Geometry whereoften the only structure present was just the metric, one gets spoiled when working in Euclidean space having so many nice properties. $\endgroup$– garyJul 15, 2017 at 1:40
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2 Answers
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Properties whose proofs rely on the triangle inequality include:
A ball is convex.
Vector addition is continuous.
answered Jul 15, 2017 at 1:25
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9$\begingroup$ @Chickenmancer How is this not addressing the question? OP asked what would happen if we relaxed this, and Andrew addressed that $\endgroup$– garyJul 15, 2017 at 1:35
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If one defines a map $\|{\cdot} \| : V\to \Bbb{R}$ that satisfies all of the conditions of being a norm, but we are less strict with the traditional triangle inequality; for example
$$\|x+y\|\leq K(\|x\|+\|y\|) $$ for some fixed $K>0,$ then we call this a quasinorm. There is much to consider of such a definition, see: http://www.ams.org/journals/bull/1945-51-01/S0002-9904-1945-08273-1/S0002-9904-1945-08273-1.pdf
When $K$ is not fixed there isn't clear answer as to what your function is actually telling you about $V$. What is desirable about the triangle inequality is that it furnishes us with a metric, that is, if you define $d(x,y)=\|x-y\|,$ then you establish $(V,d)$ as a metric space as you have already pointed out.
Others have pointed out various other conditions, all of which are contingent on the topology endowed by the metric defined by the norm. So what happens in the end when you define a map which ignores the triangle inequality, you ignore the metric topology, and in particular, you ignore any geometric aspects of your vector space which may be of use. Moreover, metric spaces can greatly reduce the difficulty of many arguments, and may even allow for analysis (depending on the field $V$ is defined over).
