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I have points that are characterized by a timestamp and a location, so 3 dimensional points, one temporal x, and two for the location coordinates.

My function $d(x, y)$ is defined as follows:

$$d(x, y) = m1 * dt(x, y) + m2 * ds(x, y),$$

where $dt$ is the temporal distance $(\text{abs(timestamp}_x - \text{timestamp}_y))$ and $ds$ is the euclidian spatial distance.

Does $d(x, y)$ conform to the triangle inequality property necessary for it to be a metric? How can I proceed to prove (or disprove) this?

That triangle inequality property for a function is: $$f(x, z) \leq f(x, y) + f(y, z)$$

I'm sorry if my question is dull or trivial, but I have a weak background in mathematics. Thanks in advance.

eat chocolate
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Prove that $dt(x,y)$ and $ds(x,y)$ are metric spaces over $\mathbb{R}$. To prove $(V,d)$, $d:V\times V\rightarrow V$ is a metric space you need to show that $d(x,x)=0,d(x,y)=d(y,x)$ and the triangle inequality $\forall x,y \in V$. Once you show that for $dt$ and $ds$, note that $d$ is a linear combination of these two and you would be able to establish these properties for $d$ as well.

  • Thanks, couldn't vote up the answer because I don't have enough reputation I'm sorry. – eat chocolate May 09 '19 at 15:33
  • But one thing I'm not sure if I understood, sorry it may be stupid, if the d is linear combination of valid metrics then d must be a valid metric as well? I'm sure dt and ds are valid metrics. – eat chocolate May 09 '19 at 15:34
  • @eatchocolate Yes. Don't be sure about anything unless you have proved it. You can prove in the same way that I outlined that their addition is also a metric space, provided $m_1,m_2$ are scalars. –  May 09 '19 at 15:42