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Math Help - Ask for two counterexamples regarding convergence in L^p

  1. #1
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    Ask for two counterexamples regarding convergence in L^p

    If <f_n> is a sequence in L^p converging in L^p to an f in L^p, that is, \|f_n-f\|_p\to 0 as n\to\infty, do we necessarily have \lim\limits_{n\to\infty}\|f_n\|_p=\|f\|_p? If not, could you please come up with a counterexample?
    If <f_n> is a sequence in L^p satisfying \lim\limits_{n\to\infty}\|f_n\|_p=\|f\|_p where f\in L^p, do we necessarily have <f_n> converges in L^p to f (as defined above)? If not, could you please come up with a counterexample?
    Thanks!
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  2. #2
    A Plied Mathematician
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    There's a decent chance you might find what you're looking for here. Maybe your library has it.
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    Quote Originally Posted by Ackbeet View Post
    There's a decent chance you might find what you're looking for here. Maybe your library has it.
    No, I don't have this book. I searched just now the contents of this book from google books and the item 40 "sequences of functions converging in different senses" might be relevant, but I didn't find any useful information regarding my question from the pages that are allowed to read by google books.
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  4. #4
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    Quote Originally Posted by zzzhhh View Post
    If <f_n> is a sequence in L^p converging in L^p to an f in L^p, that is, \|f_n-f\|_p\to 0 as n\to\infty, do we necessarily have \lim\limits_{n\to\infty}\|f_n\|_p=\|f\|_p? If not, could you please come up with a counterexample?
    Look at the reverse triangle inequality,
    <br />
|\, ||f||-||g||\, | \leq ||f-g||<br />
    If <f_n> is a sequence in L^p satisfying \lim\limits_{n\to\infty}\|f_n\|_p=\|f\|_p where f\in L^p, do we necessarily have <f_n> converges in L^p to f (as defined above)? If not, could you please come up with a counterexample?
    Thanks!
    Take the real line with the Lebesgue measure and let  f_n=\mathbf{1}_{[n,n+1]}. They have  ||f_n||_1=1 \rightarrow ||\mathbf{1}_{[0,1]}||_1 but it does not converge to that.
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    Brilliant! Thank you very much, Focus!
    So \lim\limits_{n\to\infty}\|f_n-f\|_p=0 necessarily implies \lim\limits_{n\to\infty}\|f_n\|_p=\|f\|_p, but the converse does not. Sometimes the converse may be true, sometimes may not as your counterexample indicates. Only when it is known in advance that f_n converges a.e. can the converse (convergence in L^p) necessarily hold.
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  6. #6
    A Plied Mathematician
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    Problem 6.16 in Royden, 3rd Edition seems relevant.
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    Quote Originally Posted by Ackbeet View Post
    Problem 6.16 in Royden, 3rd Edition seems relevant.
    No, they are not relevant. I didn't assume convergence a.e.
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  8. #8
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    Quote Originally Posted by zzzhhh View Post
    Brilliant! Thank you very much, Focus!
    So \lim\limits_{n\to\infty}\|f_n-f\|_p=0 necessarily implies \lim\limits_{n\to\infty}\|f_n\|_p=\|f\|_p, but the converse does not. Sometimes the converse may be true, sometimes may not as your counterexample indicates. Only when it is known in advance that f_n converges a.e. can the converse (convergence in L^p) necessarily hold.
    You need f to be in L^p in that case. More generally if the sequence is UI then a.e. convergence (or convergence in measure) implies L^p convergence. By UI I mean that
    \sup ||f_n \,\mathbf{1}_{|x|>K}|| \rightarrow 0 as  K \rightarrow \infty.
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