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Math Help - Dual Space Problem

  1. #1
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    Dual Space Problem

    Let X be a normed space and X' its dual space (set of bounded linear functionals on X).

    If X not equal to {0} show that X' not equal to {0).
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  2. #2
    AMI
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    X\neq\{0\}\Rightarrow\exists x_0\in X such that x_0\neq 0.
    Y:=\mathbb{C}x_0\subset X linear subspace, \phi:Y\to\mathbb{C},\phi(\lambda x_0):=\lambda,\forall \lambda x_0\in Y, belongs to Y^{\,\prime}.
    p:X\to[0,\infty),p(x):=\|\phi\|\cdot\|x\|,\forall x\in X is a seminorm on X with |\phi(y)|\leqslant p(y),\forall y\in Y.
    By the Hahn-Banach theorem, \exists f:X\to\mathbb{C} linear form such that f(y)=\phi(y),\forall y\in Y and |f(x)|\leqslant p(x),\forall x\in X.
    \Rightarrow|f(x)|\leqslant\|\phi\|\cdot\|x\|,\fora  ll x\in X\Rightarrow f\in X^{\,\prime}
    and f(x_0)=f(1\cdot x_0)=\phi(1\cdot x_0)=1\neq 0.
    So we found f\in X^{\,\prime} with f(x_0)\neq 0\Rightarrow f\neq 0\Longrightarrow X^{\,\prime}\neq 0.
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  3. #3
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    Great

    Great AMI thank you for your help.
    I am still working through it.
    I dont have much experience with semi-norm. What is this exactly?
    I understand Hahn-Banach Theorem.

    Much appreciated for your time and effort.


    Quote Originally Posted by AMI View Post
    X\neq\{0\}\Rightarrow\exists x_0\in X such that x_0\neq 0.
    Y:=\mathbb{C}x_0\subset X linear subspace, \phi:Y\to\mathbb{C},\phi(\lambda x_0):=\lambda,\forall \lambda x_0\in Y, belongs to Y^{\,\prime}.
    p:X\to[0,\infty),p(x):=\|\phi\|\cdot\|x\|,\forall x\in X is a seminorm on X with |\phi(y)|\leqslant p(y),\forall y\in Y.
    By the Hahn-Banach theorem, \exists f:X\to\mathbb{C} linear form such that f(y)=\phi(y),\forall y\in Y and |f(x)|\leqslant p(x),\forall x\in X.
    \Rightarrow|f(x)|\leqslant\|\phi\|\cdot\|x\|,\fora  ll x\in X\Rightarrow f\in X^{\,\prime}
    and f(x_0)=f(1\cdot x_0)=\phi(1\cdot x_0)=1\neq 0.
    So we found f\in X^{\,\prime} with f(x_0)\neq 0\Rightarrow f\neq 0\Longrightarrow X^{\,\prime}\neq 0.
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  4. #4
    AMI
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    Well, if you have X a (real or complex) linear space, then p:X\to[0,\infty) is called a seminorm if:
    1) p(x+y)\leqslant p(x)+p(y),\forall x,y\in X (i.e. p is sublinear) and
    2) p(\lambda x)=|\lambda|\cdot p(x),\forall x\in X,\forall\lambda (in \mathbb{R} or \mathbb{C}).

    The Hahn-Banach theorem in its complex form (which is a quick corollary of the initial Hahn-Banach theorem) states that if we have X a complex linear space, p a seminorm on X, Y a linear subspace of X, and g a linear form on Y such that |g(y)|\leqslant p(y),\forall y\in Y, then there exists a linear form f on X which is equal to g on Y and which satisfies the inequality |f(x)|\leqslant p(x),\forall x\in X.

    Let me know if you have any more questions
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  5. #5
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    almost there

    Thank you again that is very helpful and insightful.

    I've got a better idea now but still struggling with one thing.
    Its the way you define the mapping of Y in the second line


    <br />
    " alt="Y:=\mathbb{C}x_0\subset X
    " />

    It seems strange to me as \mathbb{C} is the symbol for the set of Complex numbers. Should this be written differently?
    Please explain this notation if you dont mind.




    Quote Originally Posted by AMI View Post
    Well, if you have X a (real or complex) linear space, then p:X\to[0,\infty) is called a seminorm if:
    1) p(x+y)\leqslant p(x)+p(y),\forall x,y\in X (i.e. p is sublinear) and
    2) p(\lambda x)=|\lambda|\cdot p(x),\forall x\in X,\forall\lambda (in \mathbb{R} or \mathbb{C}).

    The Hahn-Banach theorem in its complex form (which is a quick corollary of the initial Hahn-Banach theorem) states that if we have X a complex linear space, p a seminorm on X, Y a linear subspace of X, and g a linear form on Y such that |g(y)|\leqslant p(y),\forall y\in Y, then there exists a linear form f on X which is equal to g on Y and which satisfies the inequality |f(x)|\leqslant p(x),\forall x\in X.

    Let me know if you have any more questions
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  6. #6
    AMI
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    I defined Y to be the set \{\lambda x_0 | \lambda\in\mathbb{C}\} which can also be denoted \mathbb{C}x_0 and it can easily be shown that is a linear subspace of X.
    Is that what you were asking?!
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