1
$\begingroup$

Let $K$ be a finite set and $\mathcal{A}$ a family of functions on $K$ that is a self-adjoint algebra, separates points and vanishes nowhere. Prove that $\mathcal{A}$ must then already contain every function on $K$.

Definitions:

-Self-adjoint algebra

for $f,g\in \mathcal{A}$, $c\in \mathbb{C}$ we have $f+g\in \mathcal{A}$, $f\cdot g\in \mathcal{A}$, $cf\in \mathcal{A}$, $\bar{f}\in \mathcal{A}$.

-Separates Points

$\forall x,y\in K$ with $x\neq y$, $\exists f\in \mathcal{A}$ so that $f(x)\neq f(y)$.

-Vanishes Nowhere

$\forall x\in K$, $\exists f\in \mathcal{A}$ so that $f(x)\neq 0$.

I've tried too see if an induction argument on $K$ would work, but I've been having trouble showing that if it works for $|K|=n \Rightarrow$ works for $|K|=n+1$. And I'm beginning to believe that it won't work.

I have also tried to show that I could construct a function that is nonzero for one element and zero everywhere else. Then I could scale it and sum these functions for each element individually. But either it cannot be done or I'm not creative enough to show it.

I've been told by my professor that we need not assume any structure on $K$ ($i.e.$ we do not need to put a metric on to solve this problem) and that we need not assume the functions be complex or real valued.

$\endgroup$
  • $\begingroup$ "Every function on $K$." What kind of functions? We need to restrict the codomain. $\endgroup$ – edm Oct 20 '18 at 19:07
  • $\begingroup$ @edm It would be the complex valued functions. $\endgroup$ – G the Stackman Oct 20 '18 at 23:18
1
$\begingroup$

Without loss of generality, assume $K=\{1,2,\dots,n\}$. We want to prove that the function $f$ such that $f(1)=1$ and $f(x)=0$ elsewhere is in $\mathcal A$. The element $1\in K$ is not special in $K$. Once we prove this $f$ is in $\mathcal A$, you also show that the function that is equal to $1$ at one element and $0$ elsewhere is also in $\mathcal A$.

We want to find a function $f_2\in\mathcal A$ such that $f_2(1)\not=0$, $f_2(2)=0$. By "vanishes nowhere", find a function $g\in\mathcal A$ such that $g(1)\not=0$. If $g(2)=0$, we are done by letting $f_2=g$; if $g(2)\not=g(1)$ or $0$, consider the function $\frac{1}{g(2)}g=:h\in\mathcal A$. This function $h$ has properties

  1. $h(1)\not=0$

  2. $h(1)\not=h(2)$

  3. $h(2)=1$

  4. $h(1)\not=1$ by 2 and 3.

From these properties, you can prove that the function $hh-h\in\mathcal A$ has properties $(hh-h)(1)\not=0$ and $(hh-h)(2)=0$. To see these, note that $(hh-h)(1)=0$ implies $h(1)(h(1)-1)=0$, which the properties tell you this is not the case; while $hh(2)=h(2)=1$ so $hh(2)=h(2)=0$. So $hh-h$ is the function $f_2$ we want.

If $g(2)=g(1)$, by "separates point", find a function $k\in\mathcal A$ such that $k(1)\not=k(2)$ and $k(2)\not=0$ (if $k(2)=0$, then let $f_2:=k$). Look at the function $g-\frac{g(2)}{k(2)}k=:h_0\in\mathcal A$. This function's value at $2$ is $0$. We also have $h_0(1)\not=h_0(2)$; otherwise $$g(2)-\frac{g(2)}{k(2)}k(1)=g(1)-\frac{g(2)}{k(2)}k(1)=g(2)-\frac{g(2)}{k(2)}k(2)$$ (the first equality is by $g(2)=g(1)$, the second equality is by $h_0(1)=h_0(2)$), thus giving $k(1)=k(2)$, a contradiction. (Note that $g(2)\not=0$ so you can divide by $g(2)$) So $h_0(1)\not=0$ and $h_0(2)=0$. We let $f_2:=h_0$.

We thus found a function $f_2\in\mathcal A$ such that $f_2(1)\not=0$, $f_2(2)=0$. By the same manner, find a function $f_j\in\mathcal A$ such that $f_j(1)\not=0$, $f_j(j)=0$, for each $j\in\{2,3,\dots,n\}$.

Now look at the product $\prod_{j=2}^n f_j=:f\in\mathcal A$. This function has property $f(1)\not=0$ while $f(j)=0$ elsewhere. The function $\frac{1}{f(1)}f\in\mathcal A$ has value $1$ at $1$ and has value zero elsewhere.

$\endgroup$
0
$\begingroup$

I think it works by induction. Where does it make you feel not?

Begin with $|K| = 2$, by the property of separating points, we can get a function $f$ such that $f(x_1) \neq f(x_2)$. Then you can create two indicator functions by this and the vanishing nowhere property. Indicator functions here I mean the function is zero on everywhere except one element in $K$.

Later, you can do induction step by first creating functions that has two elements' function values nonzeros. Then, you can continue to create indicator functions.

$\endgroup$
  • $\begingroup$ I had the idea of an indictor function, where it has a non zero value at a point and 0 everywhere else, but i could not construct such a function. How do you show that this indictor function you mentioned is in the set? $\endgroup$ – G the Stackman Oct 21 '18 at 1:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.