If L(P, f) = U(P, f), prove that f is a constant function on [a, b]
Solution 1
By definition
$$m_i=\inf_{x_{i}\le x\le x_{i+1}} f(x),\quad M_i=\sup_{x_{i}\le x\le x_{i+1}} f(x)$$
Since trivially $m_i\le M_i$ and $(x_{i+1}-x_i) >0$, if one $m_i<M_i$ we have that strict inequality holds between $L(f,P)$ and $U(f,P)$, hence $m_i=M_i$ for all $i$. Then as
$$m_i\le f(x^*)\le M_i=m_i,\qquad x_i\le x^*\le x_{i+1}$$
so equality holds and $f(x^*)=m_i$. So we have that $f(x)$ is piecewise constant on each $[x_i,x_{i+1}]$ and equal to $m_i$ on these intervals for every $0\le i<n$.
But then, we have that $m_i=f(x_i)=f(x_{i+1})=m_{i+1}$, so that $m_i=m_{i+1},\; 0\le i <n$
hence all the $m_i$ are equal, and so $f$ is constant on all of $[a,b]$ since any $x\in [a,b]$ is in one of the $[x_i,x_{i+1}]$.
Solution 2
Write
$$U(P,f)-L(P,f) = \sum (M_i-m_i) (x_{i+1}-x_i) = 0 $$
Every term in the sum is nonnegative since both $(M_i-m_i) \geq 0$ and $x_{i+1}-x_i > 0$. The only way the sum evaluates to zero is if... which means $f$ is constant on each interval... finally we make sure endpoints are okay.
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mmm
Updated on November 02, 2020Comments
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mmm about 3 years
That is what I have for the beginning of my proof, but I'm not sure how to conclude that the function is constant.
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Adam Hughes almost 9 yearsMy answer assumes you're doing the sups/infs on the half open intervals, if that's inaccurate, please let me know and I can easily fix it.
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mmm almost 9 yearsWell, mi is the inf of the function on the closed interval and Mi is the sup of the function on the closed interval.
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Adam Hughes almost 9 yearsEdits complete. Easy peasy.
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Adam Hughes almost 9 yearsYou should perhaps be more explicit with "check the endpoints are OK" as that's a bit cryptic if the student doesn't know that we are looking for consistency between intervals.
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mmm almost 9 yearsthis is probably too late of me to comment and ask another question about this, but I got a little confused in your first sentence. What do you mean by "strict inequality holds between L(P, f) and U(P, f)." I guess I'm not seeing how you come to the conclusion that $m_i = M_i$ for all i. But I think I understand the rest of the proof.
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Adam Hughes almost 9 years@mmm If you have a list $a_i\le b_i$ then $\sum a_i\le\sum b_i$ if some $a_k <b_k$ then the $\le$ in the sum is also <. That's what I was using to conclude equality.
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mmm almost 9 yearsOk so you're saying that since U(p, f) is supposed to equal L(p, f), there is no way that I could have a strict inequality between any of the $m_i$'s and $M_i$'s
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Adam Hughes almost 9 years@mmm Yes, exactly.
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mmm almost 9 yearsAnd when you write x*, is this just any point in the interval between $x_i$ and $x_{i+1}$
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Adam Hughes almost 9 years@mmm Yes, that's correct