The mode minimizes the $l_0$ norm?

Question

Suppose we have a set $S$ of $N$ real numbers. Show that $$\sum_{s_i\in S}|s_i-x|^0 $$ is minimal if $x$ is equal to the mode of S.

I'm a bit confused about that, because assuming $0^0 = 1$ the whole sum should always be equal to $N$. So it looks like this sum cannot be minimised in any way.

UPDATE: So it turns out the statement of the question as written above was not correct. It was taken from a book, but in the book the statement was slightly different. It asked

Show that for $$\lim_{q->0} \sum_{s_i\in S}|s_i-x|^q $$ is minimal if $x$ is equal to the mode of S.

I believe taking the limit $q \rightarrow 0$ instead of setting $q=0$ is what makes the difference.

If you really want this expression to define the $l_0$ "norm" then you have to assume $0^0=0$, after all in the $l_0$ norm you are supposed to count the nonzeros of a vector. In this case the proposition makes more sense (but is still extremely easy to prove, essentially equivalent to the definition of mode). — Michal Adamaszek, Jul 21 '23 at 11:03
It only makes sense if you take $0^0=0$ here. My guess is this question has been asked because of analogues with the $l_1$ and $l_2$ norms. — Chris Lewis, Jul 21 '23 at 11:06

score 1 · Answer 1 · answered Jul 21 '23 at 10:05

I agree that the sum is always equal to $N.$ This doesn't mean the sum cannot be minimised in any way: it means the sum is minimised for all values of $x\in\mathbb{R}.$ Moreover, this means that the proposition in the question is true, that is, if $x$ is the mode, then $x\in\mathbb{R},$ and so $x$ does technically minimise the expression $\displaystyle\sum_{s_i\in S}|s_i-x|^0 $ because every $x\in\mathbb{R}$ minimises the expression.

It is a weird and somewhat uninteresting/trivial proposition though.

The mode minimizes the $l_0$ norm?

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