# HW 9: Challenge 2(a,b,c)

Tongzhou, 11/21

(a) Since $L(h,P)=\sum_{k=1}^n m_k(x_k-x_{k-1}),$ $m_k$ needed to be determined firstly.

By definition, $m_k =\inf\{h(x): x \in [x_{k-1},x_k]\}$.

Since $[x_{k-1},x_k]$ is any partition of $[0,1]$, $m_k =\inf \{1,2\}=1.$

Therefore, $L(h,P) = \sum_{k=1}^n m_k(x_k-x_{k-1}) =\sum_{k=1}^n 1(x_k-x_{k-1})=1(1-0) =1$.

(b) Sidework: I want to have $U(h,P) < 1+ 1/10$ which means $1(x_k-0)+2(1-x_k)<1+1/10.$ I want to determine how far $x_k$ could be away from $1$.

Thus $lim_{k \rightarrow n}(x_k-0) + 2(1-x_k)<1.1$ so that $1-x_k <0.05$.

Now choose $P= \{0,0.96,1\}$. Again I need to determine $M_k$ before I find $U(h,P)$. $M_1 = \sup\{h(0), h(0.96)\}= 1, M_2 =\sup \{h(0.96), h(1)\}=2$. $U(h,P)= 1(0.96)+2(1-0.96) =1.04 < 1.1$

(c) By the intuition of (b).

Let $\epsilon>0$, Choose $P_\epsilon = \{0, 1-\frac{\epsilon}{2},1\}$ $M_1 = \sup \{h(0),h(1-\frac{\epsilon}{2})\}=\sup \{1,1\}=1, M_2 =\sup \{h(1-\frac{\epsilon}{2}),h(1)\} =\sup \{1,2\}=2$. $U(h,P_\epsilon) = 1 (1-\frac{\epsilon}{2})+2 (1-1+\frac{\epsilon}{2})=1+\frac{\epsilon}{2}<1+\epsilon$.

#### 3 Responses

1. Jeremy LeCrone says:

https://richmond.qualtrics.com/jfe/form/SV_72SXTAOV7oVyeYB

2. Elise Favia says:

Great work. It would be helpful to the reader to state what h(x) is for a problem like this, as it is not clear what it is since the problem statement wasn’t included.

3. Gregory Bischoff says:

All good! Just a note on the notation: for ‘x_k’, the k is indexing its place in the partition. While your sum is considering the values of k up to n, remember that is only to index the left and right endpoints of an interval!

The step you include the limit on is right after you state the same fact without the limit, and your next result is just a bit of algebra so there’s no need to worry about a limit.