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Thread: A problem with limsup/liminf

  1. #1
    Moo
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    A problem with limsup/liminf

    Hi !

    This is a proof my teacher wanted us to do, but when coming to giving the solution, he got confused

    The problem is :
    $\displaystyle \left(\liminf_{n \to \infty} f_n(x)\right)^-=\limsup_{n \to \infty} \left(f_n(x)\right)^-$

    Where $\displaystyle .^-$ is 'the negative part of the function' :
    $\displaystyle f(x)=(f(x))^+-(f(x))^-$
    $\displaystyle |f(x)|=(f(x))^++(f(x))^-$
    etc...

    Also, $\displaystyle f_n$ is a sequence of measurable functions, but I'm not sure it intervenes here.
    As a sidenote, this was used in a proof of the generalization of Beppo-Levi's lemma.

    Thanks in advance !
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  2. #2
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    Quote Originally Posted by Moo View Post
    The problem is :
    $\displaystyle \left(\liminf_{n \to \infty} f_n(x)\right)^-=\limsup_{n \to \infty} \left(f_n(x)\right)^-$

    Where $\displaystyle .^-$ is 'the negative part of the function' :
    $\displaystyle f(x)=(f(x))^+-(f(x))^-$
    It will suffice to prove this for each fixed x. So fix x.

    Let $\displaystyle x_n = f_n(x)$ and $\displaystyle y_n = (f_n(x))_-$ (I prefer to use a subscript rather than a superscript to denote the "negative part"). Then $\displaystyle y_n = \begin{cases}0&\text{if }x_n\geqslant0,\\ -x_n&\text{if }x_n<0.\end{cases}$. Also, $\displaystyle \liminf_{n \to \infty} f_n(x) = \liminf_{n \to \infty}x_n := l$, and $\displaystyle \limsup_{n \to \infty} \left(f_n(x)\right)_- = \limsup_{n \to \infty}y_n$.

    If $\displaystyle l\geqslant0$ then $\displaystyle x_n\geqslant0$ for all sufficiently large n, hence $\displaystyle y_n = 0$ for all sufficiently large n, and therefore $\displaystyle \limsup_{n \to \infty}y_n = 0$. In this case, $\displaystyle l_- = 0$ also, and so $\displaystyle \limsup_{n \to \infty}y_n = \left(\liminf_{n \to \infty}x_n\right)_-$.

    Now suppose that $\displaystyle l<0$. This implies that $\displaystyle x_n<0$ infinitely often. But $\displaystyle y_n=-x_n$ whenever $\displaystyle x_n<0$, and therefore $\displaystyle \limsup_{n \to \infty}y_n = \limsup_{n \to \infty}(-x_n) = -l$. Thus in this case it is again true that $\displaystyle \limsup_{n \to \infty}y_n = \left(\liminf_{n \to \infty}x_n\right)_-$.

    I hope that argument is convincing. I found it quite hard to write it out coherently, and I'm sure that I would also have got confused if I tried to present it to a class.
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  3. #3
    Moo
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    I don't understand the idea of fixing x, because maybe we have to consider the whole function if we want to take the negative part or the liminf ? There is no doubt you're correct, but I'm just trying to understand why ><

    Can you enlighten this point please ?

    I'm sure that I would also have got confused if I tried to present it to a class.
    My teacher is less than 40 years old, it's his first year lecturing this class, and this was a serious attack to his image of "perfect teacher". I'm happier now

    Thanks a lot, as usual
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    Quote Originally Posted by Moo View Post
    I don't understand the idea of fixing x, because maybe we have to consider the whole function if we want to take the negative part or the liminf ?
    If you don't like the idea of fixing x, then simply replace $\displaystyle x_n$ by $\displaystyle f_n(x)$ and $\displaystyle y_n $ by $\displaystyle (f_n(x))_-$ throughout the proof. I was just trying to make it easier to read by simplifying the notation. Obviously I failed.

    The point is that all these operations (limsup and liminf of a sequence of functions, taking the positive or negative part) are defined pointwise. So you might as well look restrict attention to what happens at a single point x.

    Quote Originally Posted by Moo View Post
    My teacher is less than 40 years old, it's his first year lecturing this class, and this was a serious attack to his image of "perfect teacher". I'm happier now
    It has to happen to everyone sooner or later.
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  5. #5
    Moo
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    Quote Originally Posted by Opalg View Post
    If you don't like the idea of fixing x, then simply replace $\displaystyle x_n$ by $\displaystyle f_n(x)$ and $\displaystyle y_n $ by $\displaystyle (f_n(x))_-$ throughout the proof. I was just trying to make it easier to read by simplifying the notation. Obviously I failed.

    The point is that all these operations (limsup and liminf of a sequence of functions, taking the positive or negative part) are defined pointwise. So you might as well look restrict attention to what happens at a single point x.
    No, it's all clear if we omit the hypothesis of fixing x !
    The problem was if f_n is alternatively > or < 0 while x varies.
    And ould the liminf depend on x ?

    But I'm such a fool... I think there wasn't any 'x', that is he wrote f_n instead of f_n(x)
    Anyway, it's a good extension


    Also, there is some weird way I just retrieved... Would you be kind enough to tell me if it's a good start ?

    Let $\displaystyle f~:~ x \mapsto x_-$
    It's a decreasing and continuous function.

    So if I'm not mistaking, there is a property letting us say that $\displaystyle f(\liminf u_n)=\limsup f(u_n)$, but I don't know a proof of it :/



    Haha sorry, my post is just a mess as ideas come up ><



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