two little problems about limsup / liminf.

my teacher of Mathematical Analysis gave me these two problems:

1) let f be a continuous function on (0,1) and such that liminf f(x) < limsup f(x) for x that approaches 0, then for every real value L that belongs to (liminf f(x), limsup f(x)) exists {Xn} convergent sequence to 0 in (0,1) such that lim f(Xn) = L for n that approaches to infinite.

2) let f(n), g(n) be two sequences such that f(n) >= g(n) for every n natural number, is true or false that liminf f(n) >= limsup g(n) ? Give a proof of that.

i don't know how to give a proof of the first fact, about the second one i proved that is true only if the two limits exist but i didn't find a counter-exemple to justify the assumption (so it could be wrong the entire proof).

Do you get any suggestions?

Re: two little problems about limsup / liminf.

1) What does it mean that f is continuous? What happens to sequences that converge in $\displaystyle (0,1)$ when they are passed through f? So, you want to find a sequence x_{(n,1)} that converges very close to zero such that $\displaystyle f(x_{(n,1)})$ converges very close to L. Then you want a sequence of sequences $\displaystyle x_{(n,m)}$ that converges closer and closer to 0 as m and n approach infinity such that $\displaystyle f(x_{(n,m)})$ approaches closer and closer to L as m and n approach infinity. Use the results of liminf and limsup.

2) I would need to think about it.

Re: two little problems about limsup / liminf.

Quote:

Originally Posted by

**SlipEternal** 1) What does it mean that f is continuous? What happens to sequences that converge in $\displaystyle (0,1)$ when they are passed through f? So, you want to find a sequence x_{(n,1)} that converges very close to zero such that $\displaystyle f(x_{(n,1)})$ converges very close to L. Then you want a sequence of sequences $\displaystyle x_{(n,m)}$ that converges closer and closer to 0 as m and n approach infinity such that $\displaystyle f(x_{(n,m)})$ approaches closer and closer to L as m and n approach infinity. Use the results of liminf and limsup.

No, i want to prove the opposite: that for every L in (liminf f, limsup f) exists a sequence x(n) in (0,1) so that f(x(n)) = L.

If i do so i only prove that for every sequence of points in (0,1) , through f, i get a value in (liminf f, limsup f), not that i get every value in that bounded set.

I miss the opposite inclusion, in other words that (liminf f, limsup f) is included in A = {values given by f(x(n)) for all x(n) in (0,1) that goes to 0}.

P.S. I think i found one for the 2nd question: f(n) = (-1)^n + 1 and g(n) = (-1)^n, liminf f(n) = 0; limsup g(n) = 1

Re: two little problems about limsup / liminf.

Quote:

Originally Posted by

**ayrcast** No, i want to prove the opposite: that for every L in (liminf f, limsup f) exists a sequence x(n) in (0,1) so that f(x(n)) = L.

If i do so i only prove that for every sequence of points in (0,1) , through f, i get a value in (liminf f, limsup f), not that i get every value in that bounded set.

I miss the opposite inclusion, in other words that (liminf f, limsup f) is included in A = {values given by f(x(n)) for all x(n) in (0,1) that goes to 0}.

This is why you want a sequence of sequences. For any point $\displaystyle L \in \left(\liminf f, \limsup f\right)$, find a sequence of sequences such that $\displaystyle \lim_{n\to \infty} x_{(n,m)} = \dfrac{1}{m+1}$ and $\displaystyle \lim_{n \to \infty}\lim_{m\to \infty} f\left( x_{(n,m)} \right) = L$

Re: two little problems about limsup / liminf.

Another way to try problem #1 is:

For each $\displaystyle L \in \left(\liminf_{x \to 0}f(x), \limsup_{x \to 0} f(x) \right)$, let $\displaystyle A_L = \{x \in (0,1) \mid f(x)=L \}$. Show that for for every $\displaystyle \varepsilon>0$, the open ball $\displaystyle B(0;\varepsilon)\cap A_L \neq \emptyset$. Then, let $\displaystyle x_n = \sup\left( B\left(0; \dfrac{1}{n} \right) \cap A_L \right)$. Now, $\displaystyle f(x_n) \to L$ as $\displaystyle n \to \infty$.

Re: two little problems about limsup / liminf.

thanks a lot, i succeeded in proving it thanks to your last suggestion.