Is my proof of straight line being the shortest route from one point to another correct?

Question

Every proof seems to go above my head as I'm not thorough with calculus or what is being talked about in this similar question in which the author proves it using complicated terms. As a result I tried proving it on my own.

I think it has to do with triangle inequality. Can we say that any curve joint to the two points other than the straight line as sides of a polygon? Can we say that the sum of the sides of that polygon will always be more than the straight line because of triangle inequality theorem? If we can say that, it is proved much more simply than any other proof I have seen yet.

Answer 1

This is the correct idea, but there's still a lot of leg work to be done to explain why it is the 'shortest distance.' This is because the triangle inequality essentially only deals with straight lines to begin with - the quantities involved in it are norms of vectors, and vectors are essentially straight lines. This means that the triangle inequality proves that among all paths given by straight lines, the single straight line going straight from one point to the other is the best. But how do you know there isn't a really wild and wiggly curve connecting the two points? How do you define the distance along such a curve?

However, analysis furnishes us with tools to improve this proof, that you can look into. For example, if the curve connecting the two points is sufficiently nice (the term 'bounded variation', comes to mind, for example), then one can show that approximating a curve by straight lines more and more finely, and taking the limit, your proof will generalize. There are upgraded versions of the triangle inequality that deal also with integrals, which are also just called the triangle inequality.

Another approach analysis can give us to generalize this idea is in calculus of variations. On this site, there is a linked thread here which discusses this, using the Euler-Lagrange equations.

Answer 2

I think it works for polygonal paths, but this certainly does not show the result in full generality. Indeed, a different way to go, might be to define a path to be a continuous map $\lambda:[0,1] \to \mathbb R^n$.

We need a meaningful notion of "length" in this context, and if we take $d(x,y)$ to be the usual euclidian distance one can propose the following definition

The length of a curve $\alpha:[0,1] \to \mathbb R^n$ is $$\sup_{0=t_0<t_2<,\dots,<t_n=1}\sum_{i=0}^{n} d(\alpha(t_i),\alpha(t_{i+1}))$$

Indeed this makes the proof quite easy, and it is really the "limiting" analogue of your own proof:

A valid partition is given by $n=2$ where the partition is taken to be $\{t_0,t_1\}$ of the interval, in which case the definition of supremum implies that any other path is greater than the straight one.

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The more familiar definition of arc length $\int_{[0,1]}d(x(t),y(t))dt$ is equivalent to the one given for a large family of curves (and in particular class $C^1$.) See Rudin's Principles, 6.27.