Limsup and liminf

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  • #1
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Homework Statement


Let 1A stand for an indicator function and
Let fn = 1{n} and gn={1{1} n odd, 1{1} n even.
Find limn->∞sup{fn} and limsupn->∞{gn}

Homework Equations





The Attempt at a Solution


The limits are pointwise so I found given x, then
limn->∞sup{fn} = 0
limsupn->∞{gn} = {1 for x=1 or x=2 and 0 elsewhere}
Do you agree? I just wanted to check basically :)
 

Answers and Replies

  • #2
Office_Shredder
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Your definition of gn seems to suggest it is always equal to 1{1}, was this intentional?
 
  • #3
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oops no 1{1} for n odd, 1{2} for n even
 
  • #4
Ray Vickson
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oops no 1{1} for n odd, 1{2} for n even
What is fn? Your definition means that fn is a function, with fn(x) = 1 if x = n and fn(x) = 0 for x ≠ n. However, that does not seem to be what you really mean. Just tell us in words what are fn and gn---forget about trying to (mis)use indicator functions.
 
  • #5
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fn(x) = {1 if x=n, 0 elsewhere} How am I misusing indicator functions?
By definition an indicator 1A = {1 x[itex]\in[/itex] A, 0 else}
 
  • #6
jbunniii
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I think you have the right answer, but it could be stated better. First, you need to include the argument ##(x)## of the functions, and second, you should try to typeset it so it will be clearer what you mean. I assume you meant the following:
$$f_n(x) = 1_{\{n\}}(x) = \begin{cases}1 & \text{ if }x = n \\
0 & \text{ otherwise}\end{cases}$$
$$g_n(x) = \begin{cases}1_{\{1\}}(x) & \text{ if }n\text{ is even} \\
1_{\{2\}}(x) & \text{ if }n\text{ is odd}\end{cases}\text{ (for all }x\text{)}$$
And I interpreted your answers as:
$$\limsup_{n \rightarrow \infty} f_n(x) = 0 \text{ (for all }x\text{)}$$
and
$$\limsup_{n \rightarrow \infty} g_n(x) = \begin{cases}
1 & \text{ if }x = 1\text{ or }x = 2 \\
0 & \text{ otherwise} \end{cases}$$
You can right click on my equations to see how they are typeset. If you show your work, we can check whether your reasoning is right.
 
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  • #7
Ray Vickson
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fn(x) = {1 if x=n, 0 elsewhere} How am I misusing indicator functions?
By definition an indicator 1A = {1 x[itex]\in[/itex] A, 0 else}
Yes, I know what an indicator function is; that is why I wrote what I thought you meant for fn, and it turned out to be correct---that is exactly what you meant.

It is often impossible to tell when reading some messages whether or not the OP really knows what he/she is saying; often people write one thing when they mean another. (However, you could have made everything clear by saying that fn and gn are functions, and by specifying their domain.)
 
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