Extrema of the $\mathbb{R}^4 \rightarrow \mathbb{R}$ function $ f = \sqrt{ (x_i -x_j)^2 +(y_i-y_j)^2 }$

Question

I have a function $ f = \sqrt{ (x_i -x_j)^2 +(y_i-y_j)^2 }$ and I want to find the extremal points. Therefore, I calculated the gradient:

$ g= \nabla f = \frac{1}{\sqrt{(x_i -x_j)^2 +(y_i-y_j)^2}} \begin{bmatrix} x_i - x_j \\ x_j - x_i \\ y_i -y_j \\ y_j - y_i \end{bmatrix}$.

Then in defined: $ \Delta x := x_i -x_j$ and $\Delta y := y_i - y_j$. So,

$ g= \frac{1}{\sqrt{ \Delta x ^2 + \Delta y^2}} \begin{bmatrix} \Delta x \\ -\Delta x \\ \Delta y \\ -\Delta y \end{bmatrix} = 0$.

How to calculate the extrema and handle the singularity?

thanks for your help!

EDIT:

I forgot to add information about the domain: $f: \mathbb{R}^4 \rightarrow \mathbb{R}$ I also know that the minimum will occur at $\Delta x =\Delta y=0$. But how to prove this mathematically, i.e. handling the singularity?

Answer 1

The function $f$ is clearly monotonic increasing in $\Delta x^2$ and $\Delta y^2$. Therefore, the curve has only one minima, at $\Delta x = \Delta y = 0$.

Unfortunately, you can't get this from the gradient, since it doesn't exist at the origin. To see this, use a limit approach:

$$\lim_{\Delta x, \Delta y \to 0}\frac{\Delta x}{\sqrt{\Delta x^2 +\Delta y^2}}=\frac{1}{\sqrt{1 +\Delta y^2 / \Delta x^2}}$$ Note that this depends on the slope of the line we use of get to the origin, so that the limit doesn't exist.

Answer 2

This problem, if analyzed visually, describes the distance between any two points on a Cartesian $x$-$y$ plane. Since you want to find the extremas of this function, you could think of it as finding bounds to the length of the line connecting those 2 co-ordinates. Since the line can grow as large as you want it to, there are clearly no upper bounds to this. The lower bound would be at $0$ since the length cannot be a negative number. Although this isn't the proper mathematical way of solving the question, this particular technique could be used to verify the accuracy of the solution that you derive.