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Thread: Normed space

  1. #1
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    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?
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  2. #2
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    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 $y \notin C$.
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  3. #3
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    Re: Normed space

    Quote Originally Posted by SlipEternal View Post
    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?
    Quote Originally Posted by TheMR View Post
    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 $B_\epsilon(x) \subseteq C^o~?$
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    Re: Normed space

    Quote Originally Posted by Plato View Post
    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 $C^o$.
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  5. #5
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    Re: Normed space

    Quote Originally Posted by SlipEternal View Post
    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 $C$.
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  6. #6
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    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 $\{x\in X: f(x)>\sigma\lVert x\rVert\}\subseteq C^o$
    Last edited by SlipEternal; May 1st 2017 at 12:04 PM.
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  7. #7
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    Re: Normed space

    Quote Originally Posted by SlipEternal View Post
    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 $\|X\|$
    Thanks from SlipEternal
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    Re: Normed space

    Quote Originally Posted by Plato View Post
    LaTeX: \|X\| gives the norm $\|X\|$
    Very cool! Thanks!
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    Re: Normed space

    The space is X*
    Last edited by TheMR; May 2nd 2017 at 01:43 AM.
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    Re: Normed space

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

    How is the easiest way to show the equality, should I use balls: let $a\in C^{\circ}$ then $B(a,r) \subset C$...
    Last edited by TheMR; May 2nd 2017 at 12:09 PM.
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    Re: Normed space

    Quote Originally Posted by TheMR View Post
    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 $(C\setminus C_1) \cap C^o = \emptyset$.
    Thanks from TheMR
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    Re: Normed space

    int(S) is the largest open set contained in S.
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    Re: Normed space

    Is it obvious that $C_1$ is a open set that contains subsets of C?
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  15. #15
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    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 $X$.
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