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Show [itex]\phi[/itex][itex]\circ[/itex]f is Riemann integrable

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f:[a, b][itex]\rightarrow[/itex][m, M] be a Riemann integrable function and let
    [itex]\phi[/itex]:[m, M][itex]\rightarrow[/itex]R be a continuously differentable function
    such that [itex]\phi[/itex]'(t) [itex]\geq[/itex]0 [itex]\forall[/itex]t (i.e. [itex]\phi[/itex]
    is monotone increasing). Using only Reimann lemma, show that the composition [itex]\phi[/itex][itex]\circ[/itex]f is Riemann integrable.

    2. Relevant equations
    Riemann lemma - f: [a, b] [itex]\rightarrow[/itex] is Riemann integrable iff for any [itex]\epsilon[/itex]>0 [itex]\exists[/itex]a partition P such that U(P, f) - L(P, f) < [itex]\epsilon[/itex].

    Function f is Riemann integrable hence it is bounded by [m, M]. Thus [itex]\forall[/itex]
    x[itex]\in[/itex][a, b],lf(x)l [itex]\leq[/itex] max{m, M}.

    Also, since the domain of [itex]\phi[/itex] is compact and the function is monotone and increasing, by the Extreme Value Theorem, it achieves a maximum and a minimum on [m, M], hence [itex]\phi[/itex] is also bounded. Thus, [itex]\phi[/itex]((f(a)) and [itex]\phi[/itex](f(b)) is bounded by some constant, K.


    Also know since f is Riemann integrable that there exists a partition P such that
    U(P, f) - L (P, f)< [itex]\epsilon[/itex]

    We must show U(P,[itex]\phi[/itex](f(x))) - L(P, [itex]\phi[/itex](f(x)))<[itex]\epsilon[/itex].

    I think I have most of the major pieces, can someone suggest how to put it together?
    Thank you.
     
  2. jcsd
  3. Apr 23, 2012 #2

    micromass

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    Why don't you start by writing out the definition of

    [tex]U(P,f)-L(P,f)[/tex]
     
  4. Apr 24, 2012 #3
    The upper Riemann sum = (i=1[itex]\rightarrow[/itex]n) [itex]\sum[/itex]MiΔxi where Δxi=[xi-xi-1]

    The lower Riemann sum = i=(1[itex]\rightarrow[/itex]n) [itex]\sum[/itex]miΔxi where Δxi=[xi-xi-1]

    Mi=sup[f(xi): x[itex]\in[/itex][xi, xi-1]
    mi=inf[f(xi): x[itex]\in[/itex][xi, xi-1]
     
  5. Apr 24, 2012 #4

    micromass

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    OK, let [itex]c_i[/itex] (resp. [itex]d_i[/itex]) be the element of [itex][x_i,x_{i+1}][/itex], where f reaches his maximum (resp. minimum).

    We know that

    [tex]\sum_{i=1}^n (f(c_i)-f(d_i)) \Delta x_i<\varepsilon[/tex]

    Now, can you prove that [itex]\varphi\circ f[/itex] also reaches his maximum (resp. minimum) in [itex]c_i[/itex] (resp. [itex]d_i[/itex])??

    If that were true, then we have to do something with


    [tex]\sum_{i=1}^n (\varphi (f(c_i))-\varphi(f(d_i))) \Delta x_i<\varepsilon[/tex]
     
  6. Apr 24, 2012 #5
    Can't we just say that since [itex]\phi[/itex] is monotone increasing, that we know that [itex]\phi[/itex](f(ci)) (resp. [itex]\phi[/itex](f(di))) is where [itex]\phi[/itex] reaches its maximum (resp. minimum)?

    Thus, 0[itex]\leq[/itex]l [itex]\phi[/itex](f(ci) - [itex]\phi[/itex](f(di)) l [itex]\leq[/itex]2K?
     
  7. Apr 24, 2012 #6

    micromass

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    I don't see where this comes from or why it is necessary.

    But, basically, we have something of the form

    [tex]\sum_{i=1}^n {(\varphi(f(c_i))-\varphi(f(d_i)))\Delta x_i}[/tex]

    and you must associate this with

    [tex]\sum_{i=1}^n {(f(c_i)-f(d_i))\Delta x_i}[/tex]

    Do you have any result that associates [itex]\varphi(f(c_i))-\varphi(f(d_i))[/itex] with [itex]f(c_i)-f(d_i)[/itex]?? (use that [itex]\varphi[/itex] is differentiable)
     
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