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Darboux integration, show inequality

  • Thread starter xsw001
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  • #1
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Suppose f, g:[a,b]->R are bounded & g(x)<=f(x) for all x in [a,b]
for P a partition of [a,b], show that L(g,P)<=L(f,P)

I don't know whether I should show by cases since I don't know the monotonicity of the both functions f and g. It seems like that the graphs of both functions have to behave the same way from the given condition g(x)<=f(x), isn't it? Or else how would I compare g(Xi) with f(Xi-1) or g(Xi-1) with f(x) though?

Case 1) suppose both f, g are monotone increasing
Case 2) suppose both f, g are monotone decreasing
Case 3) suppose f is monotone increasing and g is monotone decreasing
Case 4) suppose f is monotone decreasing and g is monotone increasing

Proof:
Since f, g are bounded, therefore there exists upper and lower integrals.
L(g,P)=sum g(Xi-1)(change of Xi) or L(g,P)=sum g(Xi)(change of Xi)
L(f,P)=sum f(Xi-1)(change of Xi) or L(f,P)=sum f(Xi)(change of Xi)
Since change of Xi are the same, we only need to compare sum g(Xi-1) with sum f(Xi-1) or sum g(Xi) with sum f(Xi).
Since g(x)<=f(x), it follows that sum g(Xi-1) <= sum f(Xi-1) or sum g(Xi) <= sum f(Xi)
Hence L(g,P) <= L(f,P)

Does it make sense? Any comments?
 

Answers and Replies

  • #2
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Can you first say what L(f,P) is supposed to be?
 
  • #3
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L(f,P) is the Lower Darboux Sum; L(f,P)=sum mi(Xi-Xi-1)
U(f,P) is the Upper Darboux Sum; U(f,P)=sum Mi(Xi-Xi-1)
f: is the function
P: is the partition of the domain [a,b]; P={Xo, X1, ... , Xn}
mi: is the greatest lower bound [Inf(f)]
Mi: is the least upper bound [Sup(f)]
delta Xi=Xi-Xi-1: each partition interval of domain [a,b] where Xo=a and Xn=b
 
Last edited:
  • #4
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I don't think you need monotonicity of any function: take a partition

[tex] P=\{x_0,...,x_n\} [/tex]

We know that [tex] f(x)\leq g(x), \forall x\in[a,b][/tex]. So since f and g are bounded, it follows that

[tex] m_{i,f}=\inf_{x\in[x_{i-1},x_i]}{f(x)}\leq \inf_{x\in[x_{i-1},x_i]}{f(x)}=m_{i,g}[/tex]

Therefore, it is obvious that

[tex] m_{i,f}(x_i-x_{i-1})\leq m_{i,f}(x_i-x_{i-1})[/tex]

Taking sums of both sides, yield

[tex]L_{f,P}=\sum_{i=0}^n{m_{i,f}(x_i-x_{i-1})}\leq\sum_{i=0}^n{m_{i,f}(x_i-x_{i-1})}=L_{g,P)[/tex]
 
  • #5
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Yeah, I thought so too, just need second opinion. Thanks!
 

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