The minimal euclidian distance is also the minimal of any monotonic mapping of that distance? Background:
I am trying to develop a simple intuitive understanding of optimal control theory through the use of geometry. 

Setup:
Consider the distance between one "fixed point" $\bar x$ and a sequence of other points $\boldsymbol y = y_1,y_2,\dots,y_N$ in $\mathbb{R}$ e.g. the real number line. The subsequent distances is $\boldsymbol d=\sqrt{|\bar x-\boldsymbol y|^2}=|\bar x-\boldsymbol y|$. The only signal that you have available is a monotonically increasing function $f:\mathbb{R}\rightarrow\mathbb{R}$ of $d$, such that e.g. $f(d_1)\leq f(d_2)$ implies $d_1\leq d_2$. 

Questions:
1) Show that the point $y^*$ that minimises $d$ is also the distance that minimises the line integral over the monotonic transformation.
2) You're standing at some specific point $\boldsymbol y=y$ in $\mathbb{R}^2$  with distance $d=\sqrt{|\bar x-y|^2}$. Given a step size of $\epsilon<d$  you want to reduce the distance to $\bar x$ as much as possible per step. How do you formulate the optimal direction?
2.1) Given sufficiently small steps $\epsilon$ and the optimal direction after one step, show that this direction will remain optimal until the entire distance is covered.

My intuition on those are simple, but I don't know where to start the proof or formulate it rigorously.
1) Obviously the point that minimises the distance is $y = \bar x$. I dont know how to show that it minimises the integral of the monotonic transformation? My heuristics point to something like: If $d(y)\leq d(y')$ for all $\boldsymbol y$ then $f(d(y))\leq f(d(y'))$ holds for any monotonically increasing $f$.
2) I am somewhat lost here. Intuitively there are an infinite number of paths I can take, but only following the straight line between $y$ and $\bar x$ will reduce my distance maximally. How do I show this?
2.1) If you can somehow find the optimal direction in which to walk, that direction is on the straight line between the two points. E.g. you can never do any better. Also not sure how to show this.
Any calls for clarification is appreciated. 
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$This is merely an attempt to clarify the question, but it's long for a comment. Do any of the proposals below match your intent?
For notational uniformity, work in $\Reals^{n}$, and let $\Brak{\ ,\ }$ denote the Euclidean metric (the constant field of inner products whose value at each point is the standard dot product, a.k.a., "the standard flat Riemannian metric"), and $d$ the induced topological metric:
$$
d(x, y) = |x - y| = \sqrt{\Brak{x - y, x - y}}
= \inf_{\gamma} \int_{0}^{1} |\gamma'(t)|\, dt,
\tag{1}
$$
with the infimum taken over all piecewise-$C^{1}$ paths $\gamma:[0, 1] \to \Reals^{n}$ satisfying $\gamma(0) = x$ and $\gamma(1) = y$.
There are multiple ways a function $f$ can be used to "distort" distance, including:


*

*Post-composing the topological distance $d$ with a strictly monotone function $f:[0, \infty) \to \Reals$:
$$
d_{f}(x, y) = (f \circ d)(x, y) = f(d(x, y));
$$
(This interpretation is suggested by your setup.)

*Multiplying the Riemannian metric $\Brak{\ ,\ }$ by a ("position-dependent") positive, real-valued function $f$ of $n$ variables:
$$
d_{f}(x, y) = \inf_{\gamma} \int_{0}^{1} f(\gamma(t))\, |\gamma'(t)|\, dt,
$$
with the infimum taken over the same class of paths as in (1). (This interpretation is suggested by the comment, "I don't know how to show it minimizes the integral of the monotonic transformation".)

*Post-composing the Riemannian metric $\Brak{\ ,\ }$ with a strictly monotone function $f:[0, \infty) \to [0, \infty)$ satisfying $f(0) = 0$:
$$
d_{f}(x, y) = \inf_{\gamma} \int_{a}^{b} f(|\gamma'(t)|)\, dt.
$$

The function $d_{f}$ in item 1 is minimized precisely where $d$ is minimized (as you note), but if it matters, $d_{f}$ is generally not a topological metric: It fails to satisfy positive-definiteness and the triangle inequality. (If $f(u) \geq 0$ with equality if and only if $u = 0$, then $d_{f}$ is positive-definite. If $f$ is concave, i.e., if $f(u + v) \leq f(u) + f(v)$ for all non-negative $u$ and $v$, then $d_{f}$ satisfies the triangle inequality.)
The function $d_{f}$ in item 2 is another matter entirely: Its geodesics (shortest paths) are not generally line segments.
The function in item 3 is "uninteresting", because the integral is not parametrization-independent.
A: I am not sure what you mean by the integral in the first question. What are you integrating over?
For the second point: you can use that the metric is translation invariant so that it is enough to consider x=0. Then transform the metric to radial coordinates. It should then be easier to see that you decrease it maximally when you move in the $r$ direction. This then answers the third question as well, since moving a small epsilon optimally in the $r$ direction still leaves the $r$ direction the same.
edit: So the Euclidean norm that induces the metric is given by $\lvert y \rvert = \sqrt{y_1^2+y_2^2}$. If instead of the normal Euclidean coordinates for $y$ we use radial coordinates: $y = (r,\theta)$ where $r$ is the distance that $y$ is from zero and $\theta$ is the angle $y$ makes with zero, then the distance is simply $\lvert y \rvert = r$. Therefore changing the distance as much as possible means changing $r$ as much as possible. For a small vector $\epsilon$ of given length $\lvert \epsilon \rvert = \delta$ it makes $y$ the most smaller when their directions are the same, otherwise $r$ would be changed less.
