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Math Help - Verify a Proof about L2 Norm

  1. #1
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    Verify a Proof about L2 Norm

    I'm working through some suggested problems and one asks to show that if f_{n} \rightarrow f uniformly on a compact interval I, and each f_{n} continuous on I, then \displaystyle \lim_{n \rightarrow \infty} || f_{n} - f|| = 0, where ||...|| is the L^{2} norm. I first want to verify that the following is sound: If \displaystyle \lim_{n \rightarrow \infty} \int_{I} (f_{n} - f)^{2} = 0 then \displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0. My main worry is moving the limit outside of the square root in this argument.

    Anyway, given that this is sufficient, then we know that for sufficiently large n, for all x \in I |f_{n}(x) - f(x)| < \varepsilon, and since I is compact it's therefore bounded, let's say the length of the interval is bounded by M. So  \displaystyle \int_{I} |f_{n} - f|^{2} < M\varepsilon. But at no point did I seem to use continuity of f_{n}...
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  2. #2
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    Quote Originally Posted by ragnar View Post
    I'm working through some suggested problems and one asks to show that if f_{n} \rightarrow f uniformly on a compact interval I, and each f_{n} continuous on I, then \displaystyle \lim_{n \rightarrow \infty} || f_{n} - f|| = 0, where ||...|| is the L^{2} norm. I first want to verify that the following is sound: If \displaystyle \lim_{n \rightarrow \infty} \int_{I} (f_{n} - f)^{2} = 0 then \displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0. My main worry is moving the limit outside of the square root in this argument.

    Anyway, given that this is sufficient, then we know that for sufficiently large n, for all x \in I |f_{n}(x) - f(x)| < \varepsilon, and since I is compact it's therefore bounded, let's say the length of the interval is bounded by M. So  \displaystyle \int_{I} |f_{n} - f|^{2} < M\varepsilon. But at no point did I seem to use continuity of f_{n}...
    Your argumentts look fine to me. Notice we don't need I to be compact or the f_n to be continous. Assume only that \lambda (I) < \infty ( I need not be an interval) and that f_n \rightarrow f uniformly on I and your argument still works.

    Edit: Another argument that works is using Dominated convergence in L^2 to the functions |f_n|\leq |f|+\varepsilon for n\geq N(\varepsilon )
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