# Normed Space, Compactness and Transformation

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.

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

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Mathematics, Ordinary Differential Equations

systems of equations as transformations

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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 ...

#### Solution Summary

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