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Thread: Counterexample for convergence in L1

  1. #1
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    Counterexample for convergence in L1

    Suppose that $\displaystyle \{ f_n \} \subset L^1( \mu ) $ and $\displaystyle f_n $ converges uniformly to f [/tex], by a theorem that I know, $\displaystyle f \in L^1 ( \mu ) $ and $\displaystyle \int f_n \rightarrow \int f $ provided that $\displaystyle \mu (X) < \infty $, X being the domain set.

    Now, if $\displaystyle \mu (X) = \infty $, this may not be true.

    Counterexample:

    Define $\displaystyle f_n(x)= \left \{ \begin {array} {cc} \frac {1}{x} & \mbox { if }
    1 \leq x < n \\ 0 & \mbox { if } x \geq n \end {array} \right .$

    And define $\displaystyle f(x)= \left \{ \begin {array} {cc} \frac {1}{x} & \mbox { if }
    1 \leq x \\ 0 & \mbox { if } x \leq 1 \end {array} \right .$

    And $\displaystyle f_n \rightarrow f $ uniformly, but $\displaystyle f_n$ is integrable for all n while f is not.

    Question:

    Q1: Why is $\displaystyle f_n \rightarrow f$ uniformly? The book just claim it is, but I'm trying to prove it using the defintion of $\displaystyle \forall \epsilon > 0 \exists N \in \mathbb {N} $ such that $\displaystyle \mid f_n-f \mid < \epsilon $

    Q2: I understand why f is not integrable since it goes to infinity, but are $\displaystyle f_n$ integrable? Is it because $\displaystyle \int f_n = \int _1^n \frac {1}{x} $ is finite?

    Thank you!
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  2. #2
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    1) I assume your domain is $\displaystyle \mathbb{R}$ notice that to evaluate $\displaystyle \vert f_n(x)-f(x) \vert$ we need three cases:

    -) If $\displaystyle x< 1$ then $\displaystyle f_n(x)=f(x)=0$
    -) If $\displaystyle 1\leq x \leq n$ then $\displaystyle f_n(x)=f(x)=\frac{1}{x}$
    -) If $\displaystyle n<x$ then $\displaystyle \frac{1}{x}$$\displaystyle \leq \frac{1}{n}$ and $\displaystyle \vert f_n(x) -f(x) \vert = \frac{1}{x} < \epsilon$ for $\displaystyle n$ big enough.

    2) Notice that $\displaystyle f_n$ are Riemann integrable.
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