# A problem with limsup/liminf

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• Jan 8th 2009, 11:00 AM
Moo
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 (Worried)

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

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

Also, $f_n$ 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 !
• Jan 8th 2009, 12:08 PM
Opalg
Quote:

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

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

It will suffice to prove this for each fixed x. So fix x.

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

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

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

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.
• Jan 8th 2009, 12:32 PM
Moo
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 ? (Bow)

Quote:

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 (Rofl)

Thanks a lot, as usual (Wink)
• Jan 8th 2009, 01:02 PM
Opalg
Quote:

Originally Posted by Moo
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 $x_n$ by $f_n(x)$ and $y_n$ by $(f_n(x))_-$ throughout the proof. I was just trying to make it easier to read by simplifying the notation. Obviously I failed. (Crying) (Giggle)

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
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 (Rofl)

It has to happen to everyone sooner or later. (Smirk)
• Jan 8th 2009, 01:20 PM
Moo
Quote:

Originally Posted by Opalg
If you don't like the idea of fixing x, then simply replace $x_n$ by $f_n(x)$ and $y_n$ by $(f_n(x))_-$ throughout the proof. I was just trying to make it easier to read by simplifying the notation. Obviously I failed. (Crying) (Giggle)

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) (Crying)
Anyway, it's a good extension :D

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

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

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

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

Quote:

It has to happen to everyone sooner or later. (Smirk)
What was your first time ? (Rofl)