The Triangle Inequality implies that the shortest path between two points is a line segment. (in "Linear Algebra Done Right 3rd Edition" by Axler)

Question

I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.

On p.173 in this book, the author wrote as follows:

Note that the Triangle Inequality implies that the shortest path between two points is a line segment.

I have not read a proof of the above fact yet.

Does a typical proof of the above fact use the Triangle Inequality?
When we prove the above fact by calculus or something, do we need the Triangle Inequality?
Is it true that we cannot prove the above fact without the Triangle Inequality?

Answer 1

Parametrize an arbitrary path by $r(t)$ on $t\in[a,b]$ that is differentiable a.e. on $(a,b)$. Then by triangle inequality

$$\big\|r(b)-r(a) \big\|=\big\|\int_a^br'(t)dt \big\|\leq\int_a^b\big\|r'(t) \big\|dt.$$

LHS is distance of a line segment connecting two points; RHS is distance of an arbitrary path.

Answer 2

The more that I think about this question, the more I think that the answer is going to be: "By definition".

In order to determine what the shortest path $\gamma:[0,1]\to\mathbb{R}^n$ with $\gamma(0)=a$ and $\gamma(1)=b$ between two points $a$ and $b$ is, we first have to measure the length of a path. Otherwise we can not determine the shortest. So we need a way to map paths to lengths. Assuming we have such a mapping $$ \mu: ([0,1]\to\mathbb{R}^n) \to \mathbb{R}_+, $$ then it is reasonable to assume that the distance $d(a,b)$ between two points $a,b$, should be the infimum of the path lengths over all paths. I.e. $$ d(a,b) = \inf \big\{\mu(\gamma) \mid\gamma:[0,1]\to\mathbb{R}^n \text{ with } \gamma(0)=a, \gamma(1)=b \big\} $$ cf. Intrinsic metric. So by definition $\|b-a\|=d(b,a)$ should be equal to the shortest path. If the length of the straight line $\gamma_S(t) = a + t(b-a)$ is equal to $\|b-a\|$ we would be done. Now the question is, with what do we start? With a given length mapping $\mu$ for paths which has the property that its induced distance results in a well defined norm on our vector space? Or should we start with a norm and define a length mapping $\mu$ using this norm?

Starting with a norm

A reasonable definition of a path length might be, that we define straight lines to be equal to the norm. Define the length of polygonal chains to be equal to the sum of its individual line segments and then approximate paths with polygonal chains and define the length of a path to be the limit length of the approximating polygonal chains. Since any path connecting two points is therefore a sum of line segments (in the limit), it follows by the triangle inequality that the straight line connecting these two points must necessarily be smaller. But this is kind of by definition.

Starting with path lengths ($\mu$)

This is much more tricky, here we have to show that $\mu(\gamma_S)=\|b-a\|$ for the straight line $\gamma_S(t) = a + t(b-a)$. The proof is probably by contradiction: assume that this is not the case and prove that the intrinsic metric of this length measure $\mu$ does not result in a well defined norm. But I currently do not know how to do that.