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Lower and upper sums

  1. Feb 2, 2004 #1
    Which functions have the property that some upper some equals some (other) lower sum?

    Constant functions obviously do. Odd functions do in some cases. Even functions don't. Step functions won't (unless we restrict our consideration to an interval where it is constant). In fact it seems that all discontinuous functions would fail to have this property. Can anyone generalize/formalize this?
     
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  3. Feb 3, 2004 #2

    matt grime

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    upper and lower sums of what? are you talking about Riemann integration here?
     
  4. Feb 3, 2004 #3

    matt grime

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    Well, in general they will never be equal on every possible partition. Even for the function f(x)=x on the interval [0,1] that conjecture is trivially false - they differ by approximatey e.f(e) + h.f(1) where e and h are the lengths of the first and last subset in the partition.

    So, you are asking if there is always some partition, P, that has U(P)= L(P), what does that imply about f?

    Hmm, not sure. What partitions are you using for the results for some odd functions?

    In fact because if true for partition P it is true for any refinement of P, then I think it will serverely restrict potential f's
     
    Last edited: Feb 3, 2004
  5. Feb 3, 2004 #4

    NateTG

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    I assume that you mean:
    [tex]e*f(0)+h*f(1)[/tex]

    Obviously, the function [tex]f(x)=0[/tex] will always have upper and lower sums equal to [tex]0[/tex].

    To Jupiter:

    Even in more abstract situations, constant functions are continuous.

    Now, lets assume that we have a function [tex]f:\Re\rightarrow\Re[/tex] that is not constant ]on some interval [tex][a,b][/tex].
    Now, if [tex]f(a)\neq f(b)[/tex] then we have that the entire interval is a partition with differing upper and lower sums.
    Now, since [tex]f[/tex] is not constant, there is some [tex]x[/tex] such that [tex]f(x) \neq f(a)[/tex] unless [tex]x=\frac{b+a}{2}[/tex] the partition on [tex]a,x,b[/tex] will generate differing upper and lower sum, so the only remaining case is where [tex]x=\frac{b+a}{2}[/tex]
    Now, if there is some [tex]y\neqx[/tex] and [tex]f(y)\neq f(a)[/tex] then we have a partition on [tex]a,y,b[/tex] which generates differing upper and lower sums.
    Otherwise there is no such [tex]y[/tex] so for any [tex]z \in [a,b], z\neqx \rightarrow f(z)=f(a)[/tex] so we can partition along [tex]a, \frac{a+a+b}{3},x,b[/tex] and get differing upper and lower sums.

    Therefore, the only functions that can have equal upper and lower sums for all partitions are cosntant functions.
     
  6. Feb 3, 2004 #5

    matt grime

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    . and * both denote multiplication. I don't see what f(x)=0 has to do with my example f(x)=x

    secondly, the rest of your argument falls into the trap that i hit: the original question asks, about *a* partition, not *every* partition. You've shown that a 2 part partition might not do it, that doesn't say anything about other more esoteric partitions. As posed, it is not sufficient to provide some partition on which the upper and lower sums are different, but to show that on every partition they are different. (notice the word 'some' in the question.)

    I think that the refinement argument might yield something interesting.
     
    Last edited: Feb 3, 2004
  7. Feb 3, 2004 #6

    NateTG

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    Well, if you can describe the minimal and maximal U and L partitions, then you should be able to take advantange of continuity to show that for continuous functions there are patitions so that [tex]U(P_1)=L(P_2)[/tex] it's a bit tricky with non-continous functions.

    For the upper and lower values on the same partitions to be the same is a good bit trickier.
     
  8. Feb 3, 2004 #7

    matt grime

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    But you can't as the example f(x)=x shows on the interval [0,1]

    for any partition the difference is e^2+h, where e and h are the lengths of the first and last subinterval.
     
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