A Universe of Sorts

§ Minimising L2 norm with total constraint

Suppose we are trying to minimize $x^2 + y^2$ subject to $x + y = 10$ .
We can think of $(x, y)$ as two points located symmetrically about $5$ , suppose it is $x = (5 + \epsilon)$ and $y = (5 - \epsilon)$ .
See that the function $f(k) = k^2$ is such that the output becomes larger as we go to the right / increase the argument than the rate at which the output becomes smaller as we go to the left / decrease the argument.
This is clear by computing $\partial_k f = 2k$ , which means that if $k_r > k_l$ (right/left), then $\partial_{k_r} f = 2 k_r$ , while $\partial_{k_l} f = 2 k_l$ , so if we step to the left and the right by $\epsilon$ , keeping the total the same, the sum will change by $(2 k_r - 2 k_l) \epsilon > 0$ .
Said differently, because the function is convex / $f''(x) > 0$ , this means that $\partial_k|_r f > \partial_k|_l f$ , and thus we can trade the loss of the total from moving to the left (a $- \partial_k|_l \epsilon$ for the gain of the total from moving to the right (a $+ \partial_k|_r \epsilon$ ).

          * dx=1.2
         /|---->
        - |
       /  |
     --   |
*---/     |
-dx=0.8   |
  <-|     |
    |    x=0.6
   x=0.4

We gain more by moving rightwards (in terms of $f(r+dx) \simeq f(r) + f'(r) dx = f(r) + 2f(r)dx$ than we lose by moving leftward (in terms of $f(l-dx) \simeq f(l) - f'(l) dx = f(l) - 2f(l) dx$ . Since $f(r) > f(l)$ , the total we gain is still net positive.
Said differently again, we gain faster by moving from a point that is rightwards, than the rate at which we lose from a point that is leftwards.
Said differently again, the elevation gain is larger towards the right, so a small motion rightwards gains us more elevation than a small motion leftwards loses elevation.