Consider C [−1, 1]. Give examples of subspaces A ⊂ C [−1, 1] with particular properties.

54 Views Asked by At

Consider $C[−1, 1]$. Give examples (with justification) of subspaces $A \subset C[−1, 1]$ such that:

  1. $A$ is an algebra, $A$ separates points, but $A$ does not vanish nowhere.
  2. $A$ is an algebra, $A$ vanishes nowhere, but $A$ does not separate points.
  3. $A$ vanishes nowhere, $A$ separates points, but $A$ is not an algebra.

So far, I believe I have a solution for 1:

Let $A = \{f \in C[-1,1] : f(0) = 0\}$. It is easy to see that $A$ is an algebra and that it vanishes somewhere. Further, the identity function $f$ (i.e. the function defined by $f(x) = x$) is in $A$, so $A$ separates points.

And a solution for 2:

Let $A = $ Even$[-1,1]$ = $\{f \in C[-1,1] : f(-x) = f(x) \ \forall x \in [-1,1]\}$. It is clear that $A$ is an algebra as the product of two even functions is even. It does not separate points as, for example, $f(-1) = f(1)$ $\forall f \in A$. Finally, $A$ vanishes nowhere as the function $f$ given by $f(x) = 1$ $\forall x \in [-1,1]$ is in $A$.

But I cannot think of an example for 3. My first thought was to try and work with Odd$[-1,1]$ as I know that separates points and is not an algebra, but it does not vanish nowhere.

Any examples for part 3 would be appreciated.

2

There are 2 best solutions below

2
On BEST ANSWER

You mentioned the identity function $x\mapsto x$ as a candidate to make $A$ separate points, and the constant function $x\mapsto 1$ as a candidate to make $A$ vanish nowhere.

Why not try the sum of the spaces spanned by these two functions, i.e. the space of affine functions $x \mapsto ax+b$?

0
On

Your first example is the typical one for this statement. And your second example is also a good one.

For the third point you can take $A:= \{f(x) \exp(-x) : f \text{ is a polynom}\}$. This set is additive closed, but not multiplicative. Because any product is of the form $h(x) =p_1(x) p_2(x) \exp(-2x)$ and $h(x)/\exp(x)$ is not a polynomial function. It vanishes nowhere, because $(x \mapsto \exp(-x)) \in A$ and the same function can be used to show that it separates points.