# Thread: Normed space

1. ## Normed space

Let X be a real normed space, $f \in X*, \mid\mid f\mid\mid=1 \text{ and let }\sigma \in (0,1)$.
$C:=\{ x\in X: f(x)\ge \sigma\mid\mid x\mid\mid \}$.
Prove that:
$C^{\circ}=\{ x\in X\colon f(x)> \sigma \mid\mid x \mid \mid \}$.

Can you help me with this one?

2. ## Re: Normed space

Consider $C_1 = \{x\in X: f(x) > \sigma \lVert x\rVert \}, C_2 = \{x\in X: f(x) = \sigma \lVert x\rVert \}$. By definition, $C_1\cup C_2 = C$ and $C_1 \cap C_2 = \emptyset$.

So, show that for any $x \in C_1$, there exists $\epsilon>0$ such that $B_\epsilon(x) \subseteq C$.

Then, show that for any $x \in C_2$, for all $\epsilon>0$ there exists $y \in B_\epsilon(x)$ such that $y \notin C$.

3. ## Re: Normed space

Originally Posted by SlipEternal
Consider $C_1 = \{x\in X: f(x) > \sigma \lVert x\rVert \}, C_2 = \{x\in X: f(x) = \sigma \lVert x\rVert \}$. By definition, $C_1\cup C_2 = C$ and $C_1 \cap C_2 = \emptyset$.
So, show that for any $x \in C_1$, there exists $\epsilon>0$ such that $B_\epsilon(x) \subseteq C$.
Then, show that for any $x \in C_2$, for all $\epsilon>0$ there exists $y \in B_\epsilon(x)$ such that $y \notin C$.
I have a question about notation?
Originally Posted by TheMR
Let X be a real normed space, $f \in X*, \mid\mid f\mid\mid=1 \text{ and let }\sigma \in (0,1)$.
$C:=\{ x\in X: f(x)\ge \sigma\mid\mid x\mid\mid \}$.
Prove that:
$C^{\circ}=\{ x\in X\colon f(x)> \sigma \mid\mid x \mid \mid \}$.
Can you help me with this one?
In the O.P. $C^o$ is the same as SlipEternal's $C_1$.
Notations vary widely. But $C^o$ is generally use to denote the interior of a set $C$.
If that is the case here is it not enough to show that
for any $x \in C^o$, there exists $\epsilon>0$ such that $B_\epsilon(x) \subseteq C^o~?$

4. ## Re: Normed space

Originally Posted by Plato
I have a question about notation?

In the O.P. $C^o$ is the same as SlipEternal's $C_1$.
Notations vary widely. But $C^o$ is generally use to denote the interior of a set $C$.
If that is the case here is it not enough to show that
for any $x \in C^o$, there exists $\epsilon>0$ such that $B_\epsilon(x) \subseteq C^o~?$
I am saying that he should prove that $C_1 = C^o$. By definition, for any $x \in C^o$, there exists $\epsilon>0$ such that $B_\epsilon(x) \subseteq C^o$. What he is asked to prove is that the set on the RHS is equal to $C^o$. I renamed that set $C_1$. What he must prove is that for any $x\in C_2$ (which we know to be the boundary of $C$, but have not yet proven it), it does not also belong in $C^o$.

5. ## Re: Normed space

Originally Posted by SlipEternal
I am saying that he should prove that $C_1 = C^o$. By definition, for any $x \in C^o$, there exists $\epsilon>0$ such that $B_\epsilon(x) \subseteq C^o$. What he is asked to prove is that the set on the RHS is equal to $C^o$. I renamed that set $C_1$. What he must prove is that for any $x\in C_2$ (which we know to be the boundary of $C$, but have not yet proven it), it does not also belong in $C^o$.
I just do not read it that way.
I agree that "What he is asked to prove is that the set on the RHS is equal to $C^o$."
I disagree with "By definition, for any $x \in C^o$, there exists $\epsilon>0$ such that $B_\epsilon(x) \subseteq C^o$.

Now by definition if "$\bf {C_2}$ (which we know to be the boundary of $C$," then any open containing a point of $C_2$ must also a point of $C$ and a point not in $C$.

6. ## Re: Normed space

Perhaps I should rephrase. To prove equality, the OP need $C^o \subseteq \{x\in X: f(x)>\sigma\lVert x \rVert\}$ and $\{x\in X: f(x)>\sigma\lVert x\rVert\}\subseteq C^o$

7. ## Re: Normed space

Originally Posted by SlipEternal
Perhaps I should rephrase. To prove equality, the OP need $C^o \subseteq \{x\in X: f(x)>\sigma\lVert x \rVert\}$ and $\{x\in X: f(x)>\sigma\lVert x\rVert\}\subseteq C^o$
Now we agree. I, for one, need to know more about the context of this question. Knowing that in functional analysis notions vary widely. But I like to know more about the space $X^*$ or is it $X_{*}~?$

LaTeX: \|X\| gives the norm $\|X\|$

8. ## Re: Normed space

Originally Posted by Plato
LaTeX: \|X\| gives the norm $\|X\|$
Very cool! Thanks!

9. ## Re: Normed space

The space is X*

10. ## Re: Normed space

It's a dual space for X. X*:=L(X,\K), where L(X,Y) is linear space.

11. ## Re: Normed space

How is the easiest way to show the equality, should I use balls: let $a\in C^{\circ}$ then $B(a,r) \subset C$...

12. ## Re: Normed space

Originally Posted by TheMR
How is the easiest way to show the equality, should I use balls: let $a\in C^{\circ}$ then $B(a,r) \subset C$...
What are you using for the definition of the interior of a set? I have always used that it is the union of all subsets of a set which are open in $X$ (or whatever your superset may be). If that is the case, you want to show $C^o \subseteq C_1$ and $C_1\subseteq C^o$ (using my $C_1$ from above). Alternately, you can show that $C_1 \subseteq C^o$ and $(C\setminus C_1) \cap C^o = \emptyset$.

13. ## Re: Normed space

int(S) is the largest open set contained in S.

14. ## Re: Normed space

Is it obvious that $C_1$ is a open set that contains subsets of C?

15. ## Re: Normed space

It is obvious that $C_1$ is an open subset of $C$, but if you are trying to show that it is open, you need to show that it is an open subset of $X$.