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    Normed Space, Compactness and Transformation

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    Let X be a normed space, I closed interval ( or half-open on the right) and
    a = inf I, b = sup I.

    Let h : I -> [0,infinity) be a continuous function such that

    integral ( from a to b ) h(t)dt < positive infinity

    where integral from a to b represents the improper integral when I is not closed.

    Let epsilon > 0 and X_epsilon,h be the set of all continuous functions f : I -> X such that the number ||f||_e,h defined by

    ||f||_epsilon,h = (for t in I) sup( e^-(epsilon*integral(from a to t ) h(s)ds)) ||f(t)|| is finite.

    Prove that the when the interval I is compact, the set X defined above of normed spaces coincides with C(I,X), and that the sup norm and the norm ||.||_epsilon,h are equivalent. That is, Lim_n x_n = x holds in X_epsilon,h if and only if
    lim_n x_n = x uniformly.

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    https://brainmass.com/math/integrals/normed-space-compactness-transformation-50884

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    Mathematics, Ordinary Differential Equations

    systems of equations as transformations
    ________________________________________
    Let X be a normed space, I closed interval ( or half-open on the right) and
    a = inf I, b = sup I.

    Let h : I -> [0,infinity) be a continuous function such that

    integral ( from a to b ) h(t)dt < positive infinity

    where integral from a to b represents the improper integral when I is not closed.

    Let epsilon > 0 and X_epsilon,h be the set of all continuous functions f : I -> X such that the number ||f||_e,h defined by

    ||f||_epsilon,h = (for t in I) sup( e^-(epsilon*integral(from a to t ) h(s)ds)) ||f(t)|| is finite.

    Prove that the when the interval I is compact, the set X defined ...

    Solution Summary

    Normed Space, Compactness and Transformation are investigated. The solution is detailed and well presented.

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