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Math Help - Banach spaces or not?

  1. #1
    Senior Member bkarpuz's Avatar
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    Question Banach spaces or not?

    Dear MHF members,

    I have two questions on Banach spaces,
    and I would really appreciate if you help me with a complete answer.
    Let [a,b] be a interval and c\in[a,b].
    1. Denote by \Omega_{c}([a,b],\mathbb{R}) the set of all real valued bounded functions on [a,b],
      which may have jump type discontinuity from the left-hand side at the point c,
      i.e., \lim_{t\to c^{-}}f(t)\neq f(c) and \lim_{t\to c^{+}}f(t)=f(c).
      Is \Omega_{c}([a,b],\mathbb{R}) a Banach space with the \sup norm?
    2. Denote by \Psi_{c}([a,b],\mathbb{R}) the set of all real valued bounded functions on [a,b],
      which may have removable type discontinuity at the point c.
      Is \Psi_{c}([a,b],\mathbb{R}) a Banach space with the \sup norm?

    Thanks.
    bkarpuz
    Last edited by bkarpuz; March 31st 2011 at 02:21 PM. Reason: Tried to clear the definitions.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Dear MHF members,

    I have two questions on Babach spaces,
    and I would really appreciate if you help me with a rigorous answer.
    Let [a,b] be a interval and c\in[a,b].
    1. Denote by \Omega_{c}([a,b],\mathbb{R}) the set of all real valued bounded functions on [a,b], which might have jump type discontinuity from the left-hand side at the point c, i.e., \lim_{t\to c^{-}}f(t)\neq f(c) and \lim_{t\to c^{+}}f(t)=f(c).
      Is \Omega_{c}([a,b],\mathbb{R}) a Banach space with the \sup norm?
    2. Denote by \Psi_{c}([a,b],\mathbb{R}) the set of all real valued bounded functions on [a,b], which might have removable type discontinuity at the point c.
      Is \Psi_{c}([a,b],\mathbb{R}) a Banach space with the \sup norm?

    Thanks.
    bkarpuz
    What about \displaystyle f_n(x)=\frac{\mathbf{1}_{[c,b)}}{n}. Then it's clear that \left\{f_n\right\} is a Cauchy sequence in \Omega_c([a,b]) yet f_n(x)\to 0
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Drexel28 View Post
    What about \displaystyle f_n(x)=\frac{\mathbf{1}_{[c,b)}}{n}. Then it's clear that \left\{f_n\right\} is a Cauchy sequence in \Omega_c([a,b]) yet f_n(x)\to 0
    Forgive me because I dont understand it, what does this mean exactly?
    The zero function is in both of the spaces \Omega and \Psi.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Forgive me because I dont understand it, but what does this mean exactly?
    The zero function is in both of the spaces \Omega and \Psi.
    So \Omega_n([a,b]) is the space of all functions which may have a left-hand discontinuity at c??? So, in other words it's all bounded functions?
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  5. #5
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Drexel28 View Post
    So \Omega_c([a,b]) is the space of all functions which may have a left-hand discontinuity at c?
    Yes, and c is a given fixed point in [a,b].
    Last edited by bkarpuz; March 30th 2011 at 06:02 PM.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Yes, and c is a fixed point in [a,b].
    I'm just misunderstanding the point of the extra condition. Namely, if the function needn''t have a left-hand discontinuity there, why even mention it? That's like saying L(\mathbb{R}) is the set of all functions which may have infinite integral....i.e. all functions.
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I'm just misunderstanding the point of the extra condition. Namely, if the function needn''t have a left-hand discontinuity there,
    why even mention it? That's like saying L(\mathbb{R}) is the set of all functions which may have infinite integral....i.e. all functions.
    Just to have a space including the continuous functions together with some new functions which have discontinuity at a given point.
    Lets say the continuous functions on [0,1] and those which have left-hand dc at 1/2.
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  8. #8
    Senior Member bkarpuz's Avatar
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    Arrow

    Here are my arguments for the first question.
    I will try to give an affirmative answer.

    Proof of 1. It is clear that \|\cdot\|:=\sup is a norm on \Omega_{c}([a,b],\mathbb{R}).
    To complete the proof, we need to show that any Cauchy sequence \{f_{k}\}_{k\in\mathbb{N}}\in\Omega_{c}([a,b],\mathbb{R})
    converges a function in this space.
    Then, for every, \varepsilon>0, there exists k_{0}\in\mathbb{N} such that \|f_{n}-f_{m}\|<\varepsilon/2 for all m,n\in\mathbb{N} with m,n\geq k_{0}.
    In particular, for each x\in[a,b], we have |f_{n}(x)-f_{m}(x)|<\varepsilon/2 for all n,m\in\mathbb{N} with m,n\geq k_{0},
    i.e., \{f_{k}(x)\}_{k\in\mathbb{N}}\subset\mathbb{R} is a Cauchy sequence, and thus for each x\in[a,b],
    the limit f(x):=\lim_{k\to\infty}f_{k}(x) exists, to complete the proof, we have to now show that
    \{f_{k}\}_{k\in\mathbb{N}} converges uniformly to f and f\in\Omega_{c}([a,b],\mathbb{R}).

    We have |f_{n}(x)-f_{m}(x)|<\varepsilon/2 for all x\in[a,b] and all m,n\in\mathbb{N} with m,n\geq k_{0},
    which yields by letting m\to\infty that |f_{n}(x)-f(x)|<\varepsilon for all x\in[a,b] and all n\in\mathbb{N} with n\geq k_{0},
    and proves that \{f_{k}\}_{k\in\mathbb{N}} converges uniformly to f.

    Let x_{0}\in[a,b], then for every \varepsilon>0, there exists n_{0}\in\mathbb{N} such that |f(x_{0})-f_{k}(x_{0})|<\varepsilon/3 for all k\geq n_{0}.

    Let x_{0}\neq c. As the functions are continuous at x_{0}, there exists a neighborhood U_{n_{0}} of x_{0}
    such that |f_{n_{0}}(x)-f_{n_{0}}(x_{0})|<\varepsilon/3 for all x\in U_{n_{0}}.
    Then, for any x\in U_{n_{0}}, we have
    |f(x)-f(x_{0})|=|f(x)-f_{n_{0}}(x)+f_{n_{0}}(x)-f_{n_{0}}(x_{0})+f_{n_{0}}(x_{0})-f(x_{0})|
    ......................... \leq|f(x)-f_{n_{0}}(x)|+|f_{n_{0}}(x)-f_{n_{0}}(x_{0})|+|f_{n_{0}}(x_{0})-f(x_{0})|
    ......................... \leq\dfrac{\varepsilon}{3}+\dfrac{\varepsilon}{3}+  \dfrac{\varepsilon}{3}=\varepsilon
    for all x\in U_{n_{0}} proving that f is continuous at x_{0}\neq c.

    Let x_{0}=c, then we follow similar steps above with a right neighborhood of x_{0},
    and show that f is right-continuous at x_{0}=c.

    This shows that f\in\Omega_{c}([a,b],\mathbb{R}), and thus it is complete.
    Hence, \Omega_{c}([a,b],\mathbb{R}) is a Banach space. \rule{0.2cm}{0.2cm}

    It would be very appreciated if you confirm that my arguments are correct.
    Last edited by bkarpuz; March 31st 2011 at 04:34 PM.
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  9. #9
    MHF Contributor
    Opalg's Avatar
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    That argument looks correct to me.
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