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Suppose that we have $$\mathbf{a}_i - \mathbf{a}_j = \mathbf{b} + \mathbf{c},$$ where all variables are of the same dimension (i.e., $n$ dimensional). If we let $g(\mathbf{a}_i-\mathbf{a}_j)$ denote the squared Euclidean distance between $\mathbf{a}_i$ and $\mathbf{a}_j$, then the following is not true $$g(\mathbf{b}+\mathbf{c}) = g(\mathbf{b}) + g(\mathbf{c})\;$$ because $g(\mathbf{b}+\mathbf{c})=(b_1+c_1)^2+...+(b_n+c_n)^2 \ne b_1^2+c_1^2+...+b_n^2+c_n^2=g(\mathbf{b}) + g(\mathbf{c}).$

Is there any distance function $g(\cdot)$ such that $$g(\mathbf{b}+\mathbf{c}) = g(\mathbf{b}) + g(\mathbf{c})\;?$$

Ron Snow
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    What are you defining as distance function? Metrics induced by norms? Norms themselves? – gist076923 Feb 24 '23 at 19:30
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    It's a little odd call $|\cdot|$ a "distance" function. A distance function (usually called a metric) is a function of two arguments (the two objects you are measuring how far they are). – jjagmath Feb 24 '23 at 19:33
  • The question in the header doesn't match the one in the body (and neither question is clear). – lulu Feb 24 '23 at 19:37
  • @gist076923 I updated the question's notation. – Ron Snow Feb 24 '23 at 19:39
  • @jjagmath I updated the question's notation. – Ron Snow Feb 24 '23 at 19:39
  • @lulu Is it more clear now? – Ron Snow Feb 24 '23 at 19:40
  • A trivial $g$ is $g(\mathbf v) = 0$ for all $\mathbf v$. – peterwhy Feb 24 '23 at 19:40
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    No, not really. As others have remarked, a "distance function" needs two arguments, not $1$. Maybe you are talking about a norm, but that's not clear either. I suggest: rather than using words loosely, why not just list the properties you want your function to have? – lulu Feb 24 '23 at 19:41
  • @lulu I suppose g() is inspired by the squared Euclidean distance function, but it is not a distance function itself. – Ron Snow Feb 24 '23 at 19:42
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    You need to clarify what you mean by a "distance function". If you mean a norm (or some function of a norm like the squared euclidean distance), then this is trivially doesn't exist because homogeneity scuppers things via $g(0) = g(x) + g(- x) = 2g(x)$. To be more formal, look at, e.g., the properties in the standard definitions of a norm (or metric). Which of these do you want to retain? – stochasticboy321 Feb 24 '23 at 19:42
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    "inspired by" is not precise. Just list the exact properties you have in mind. – lulu Feb 24 '23 at 19:43
  • @lulu I think the answers cover the issue. Thanks for commenting, though! – Ron Snow Feb 24 '23 at 19:45
  • @stochasticboy321 Based on your comment and the answers, it looks like what I want is not going to be possible. Thanks for commenting! – Ron Snow Feb 24 '23 at 19:46
  • When you rewrote the $|\cdot|$ as $g(\cdot)$, do you still want to preserve that $g(\cdot) \ge 0$, like other "distance functions" or norms? – peterwhy Feb 24 '23 at 19:52
  • Again, you should stop calling "distance" a function of only one argument. "The distance of Chicago" makes no sense. "The distance between Chicago and New York" makes sense. See? You need to talk about the distance between two things. You could call the function the "size" of the vector or the "norm" of the vector. – jjagmath Feb 24 '23 at 21:57
  • @peterwhy Yes, that would be preferable! :) – Ron Snow Feb 25 '23 at 16:26
  • @jjagmath Great point. The function of g should be something like "size" rather than distance. – Ron Snow Feb 25 '23 at 16:26

2 Answers2

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If you are considering a normed space $(X,\|\cdot\|)$ and $X$ consists of more than $0$, then $\|x+y\| = \|x\| + \|y\|$ cannot hold for every $x,y$. Take for example $x\neq y$ and $y = -x$, then you get $0 = \|x-x\| = \|x\| + \|x\|$, which implies $\|x\| = 0 \Leftrightarrow x = 0$, which is a contradiction.

chirico
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Actually, if $X$ is any set with at least $2$ elements then you can't define any metric on it such that the triangle inequality is always an equality. Indeed, if $x\ne y$ then $d(x,x)=0$ is strictly smaller than $d(x,y)+d(y,x)=2d(x,y)$.

In particular, a norm like you want exists only for a trivial vector space.

Mark
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