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.
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.