Intuitive explanation of p-norm in finite and infinite dimensinos

Question

I am not a mathematician, so very rigorous treatment with things that only a math major learns will not suffice here.
I want to learn about p-norms and i can't quite get the intuition behind them. I am talking about what they actually mean and what their (geometric or not) consequences are.
For example, in this wikipedia page: https://en.wikipedia.org/wiki/Lp_space#The_p-norm_in_finite_dimensions
I can't undestand why for example the unit circle in L1 is a rhombus and why is it a square for p=infinity. I think not understanding these has to do with the fact that i have no intuition regarding the p-norms and the geometric consequences of the way that we defined them. And lastly, why would someone define the norm in a way that p is not equal to 2?
Thank you!

Answer 1

For the form of the sphere for $l^p$ norms, well, it's not really conceptual, you just look at the definition.

For the $l^1$ in $\mathbb{R}^2$ for instance, $(x,y)$ is a unit vector iff $|x| + |y| = 1$. If $x$ and $y$ are positive, this means $y=1-x$, so the intersection of the unit sphere with the first quadrant is the line $y= 1-x$. You can check the other quadrants to see that in total it makes a square.

Same thing for $l ^\infty$: $(x,y)$ is a unit vector iff $\max(|x|, |y|) = 1$, which means that $x=\pm 1$ and $ -1 \leqslant y\leqslant 1$, or $y=\pm 1$ and $ -1 \leqslant x\leqslant 1$ : this is also a square.

As for why someone would define these norms, I guess the answer will not be very satisfactory in finite dimension, since it is well-known that all the norms are equivalent, so indeed you can't gain much by considering all these "exotic" norms (they are not very exotic, though).

But first, some problems naturally involve powers that are not squares, and in that case the $l^p$ norm may be more natural to solve the problem.

Second and more important, these are baby cases for the real $l^p$ norms, those in infinite dimension. And they are very important, because this time they are not equivalent, so they don't give the same answers. Some objects will have a finite $l^p$ norm for some $p$ and some will not, so you have to choose your $p$ according to the situation, and $p=2$ will not suffice.