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Homework Help: Limsup and liminf

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data
    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}
    2. Relevant equations

    3. 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 :)
  2. jcsd
  3. Sep 26, 2013 #2


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    Your definition of gn seems to suggest it is always equal to 1{1}, was this intentional?
  4. Sep 26, 2013 #3
    oops no 1{1} for n odd, 1{2} for n even
  5. Sep 26, 2013 #4

    Ray Vickson

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    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.
  6. Sep 26, 2013 #5
    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}
  7. Sep 26, 2013 #6


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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:
    $$\limsup_{n \rightarrow \infty} f_n(x) = 0 \text{ (for all }x\text{)}$$
    $$\limsup_{n \rightarrow \infty} g_n(x) = \begin{cases}
    1 & \text{ if }x = 1\text{ or }x = 2 \\
    0 & \text{ otherwise} \end{cases}$$
    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.
    Last edited: Sep 26, 2013
  8. Sep 26, 2013 #7

    Ray Vickson

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    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.)
    Last edited: Sep 26, 2013
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