show that $U(P_n,f)\ge U(P_{n+1},f)$

Question

Let $f:[0,1]\to R$, and $P_n=\{0, \frac{1}{n}, ... , \frac{n-1}{n},1\}$ be a partition on $[0,1].$ Then, we can define the sequence of upper Darboux sum as $\{U(P_n,f)\}_{n=1}^{\infty}$ . I think it is true that $U(P_n,f)\ge U(P_{n+1},f)$ for $\forall n \in Z^+$, but how can I prove this? Could you give some hint?

Answer 1

For fixed $\alpha\in[0,1]$ let $f(x)=\begin{cases}1,&x<\alpha\\0,&x\ge \alpha\end{cases}$. Then clearly $U(P_n,f)\to \alpha$, but in general this convergence is not eventually monotonic. In fact, we have$$\tag1U(P_n,f)=\frac 1n(1+\lfloor n\alpha\rfloor).$$ Now consider $$\alpha=\frac7{18}$$ By induction, one sees that $10^m\cdot 7\equiv 16\pmod{18}$ for $m\ge 1$. Therefore, if $n=10^m$ with $m\ge 1$, we have $$ n\alpha-\lfloor n\alpha\rfloor = \frac {16}{18}=\frac 89$$ and $$ (n+1)\alpha-\lfloor (n+1)\alpha\rfloor = \frac 5{18}. $$ So for such $n$, we have $$ U(P_n,f)=\frac{1+\lfloor n\alpha\rfloor}n = \alpha+\frac{1}{9n}$$ and $$ U(P_{n+1},f)=\frac{1+\lfloor (n+1)\alpha\rfloor}{n+1} = \alpha+\frac{13}{18(n+1)}.$$ Thus whenever $n$ is a power of $10$, we have $$ U(P_{n+1},f)- U(P_{n},f)=\frac{13}{18(n+1)}-\frac{1}{9n}=\frac{11n-2}{18n(n+1)}>0.$$

Answer 2

Here's a way to think about it. Let's say $f$ has a sup on some interval $[a,b]$. Now, let's split the interval in two intervals, $[a,b] = [a,b_1] \cup [b_1,b]$. Now, the sup on $[a,b]$ is greater than or equal to the sup on $[a,b_1]$ and the sup on $[b_1,b]$. This is because when you split the interval, you introduce the possibility that the old max is not contained in one of the two intervals. However, at the same time, because the two new intervals combine to form the old interval, the total interval length stays the same, while the sup decreases.