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Math Help - Norms and Banch spaces

  1. #1
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    Norms and Banch spaces

    1. Show that two norms are equivalent iff the convergence of a sequence in the first norm is equivalent to its convergence in the second norm.

    2. Let C^1[0,1] be the space of all real valued functions on [0,1] which have continuous derivatives and define the norm by:

    ||f|| = f(0) + sup|f'(t)| .

    Is it a Banach spae.

    For 1. I can show the ---> implication. If they're equivalent there exists M,m > 0: 0 \leq m || x_n - x || \leq || x_n - x||' \leq M ||x_n - x || so if the sequence (x_n) converges in ||.|| then it must converge in ||.||'. Analogously for the other way around.
    For 2. I know that C[0,1] is complete with the Supremum norm. Since f' is cts. we can rewrite this as:

    ||f|| = f(0) + ||g||_{\infty} where f'(t) = g (t). So the new norm differs from the sup norm by a constant and then somehow show that they are equivalent. Then using a lemma: two norms are equivalent iff the normed spaces associated with each are both Banach finish it off.
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  2. #2
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    Quote Originally Posted by davidmccormick View Post
    2. Let C^1[0,1] be the space of all real valued functions on [0,1] which have continuous derivatives and define the norm by:

    ||f|| = f(0) + sup|f'(t)| .

    Is it a Banach space.
    You need to show that every Cauchy sequence converges in this norm. So let (f_n) be Cauchy. Then \|f_m-f_n\| = |f_m(0) - f_n(0)| + \|f_m'-f_n'\|_\infty \to0 as m,n\to\infty. But if the sum of two positive quantities tends to 0 then so do both of them individually. Therefore (f_n(0)) is Cauchy in \mathbb{R}, and (f_n') is Cauchy in C[0,1] (with the sup norm). Since both those spaces are complete, there exist a real number c and a continuous function g such that f_n(0)\to c and f_n'\to g.

    Now define f(x) = c + \int_0^x\!\!\!g(t)\,dt and show that f_n\to f in the given norm.
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  3. #3
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    Sorry I follow Opalg's idea in some sort of way....but how do you show f_n tends to f?

    (feel really stupid for asking this!)
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  4. #4
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    Quote Originally Posted by vinnie100 View Post
    Sorry I follow Opalg's idea in some sort of way....but how do you show f_n tends to f?
    Use the fact that f_n(x) - f_n(0) = \int_0^x\!\!\!f_n'(t)\,dt (fundamental theorem of calculus). Then f(x) - f_n(x) = c-f_n(0) + \int_0^x\!\!\!(g(t)-f_n'(t))\,dt.
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  5. #5
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    Wow that makes perfect sense now and now I see why it is a Banach Space
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