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Given $$ f(x)= \begin{cases} 1/x,& 0< x\le 1\\ 0,& x=0 \end{cases} $$ and a partition $P = [0, x_1, x_2,\dots, x_{n-1},1]$ and $I_k = [x_{i-1}, x_i]$, e.g. $I_1 = [0, x_1]$. Is it possible to compute the supremum and infimum of $M_k = \underset{I_k}{\mathbb{sup}} f(x)$ and $m_k = \underset{I_k}{\mathbb{inf}} f(x)$?

Any help is highly appreciated.

darkmoor
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3 Answers3

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For $k\ge 2$, you have that $I_k$ is a closed interval and $f(x)=1/x$ is a continuous function on it, so it attains its $\min$ and its $\max$ on this interval (which Theorem is it?). Hence, for $k\ge 2$ \begin{align}\sup_{I_k}f(x)=\max_{I_k}f(x)&=f(x_{k-1})\\ \inf_{I_k}f(x)=\min_{I_k}f(x)&=f(x_{k})\end{align} Now, in $I_1=[0,x_1]$, you have that $\lim_{x\to 0} f(x)=+\infty$, so $$\sup_{I_1}f(x)=+\infty$$ On the other hand $f(0)=0$, and $f(x)=1/x>0$ for any other $x$, so $$\inf_{I_1}f(x)=0$$

Edit: To elaborate on $I_1$: $f$ has exactly one value on $x=0$, which is $f(0)=0$. However, $\lim_{x\to 0+} f(x)=+\infty$, which is different than $f(0)$. This does not mean that $f$ has two values for $x=0$. It means that $f$ is discontinuous at $x=0$, since $$0=f(0)\neq \lim_{x\to0+}f(x)=+\infty$$ Now, as $x$ tends to $0$, $f(x)$ increases all the time without bounds. In other words, name a number, then there is an $x$ (very small, meaning very close to $0$) so that $f(x)=1/x$ is bigger than this number you named. This shows that $$\sup_{I_1}f(x)=\lim_{x\to 0}f(x)=+\infty$$ Note: here you have indeed a $\sup$ and not a maximum (since it is attained asymptotically). Now, for the $\inf$ things are simpler. $f$ takes the value $0$ (exactly, not asysmptotically, or anything the like, but exactly), at $x=0$ and is positive for any other value in $I_1$. Hence, $f(0)=0$ is indeed the least value of $f$ in $I_1$, which establishes that $$\inf_{I_1}f(x)=\min_{I_1}f(x)=f(0)=0$$

Jimmy R.
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  • thanks for the answer. I understand clearly the case where $k>1$ but I found the $I_1$ case a litle obscure. I am trying to understand it step by step. That is, the first step is to find the lower and upper bounds of the range of $f$ in $I_1$ and in second step is to peak the $\mathbb{sup} and $\mathbb{inf}. This is where obscurity begins. I am not sure what is the problem, maybe because $f$ has two values in $0$. Could you please elaborate a litle more in this point? I believe an example following the abovementioned steps will be great. Thank you again! – darkmoor Aug 11 '17 at 19:00
  • I think I got it in a more analytical form. Given, $C={f(x)| x\in[0,x_1]}={f(x) | x\in(0,x_1]} \cup {f(x)|x=0}={f(x) \in [1/x_1, +\infty)} \cup {0}$, we may take, $\sup C = \sup \big({ f(x) \mid x \in [1/x_1, +\infty)} \cup {0}\big) = \sup\big{ \sup{f(x) \in [1/x_1, \infty)},\sup{0} \big} = \sup {+\infty, 0}= +\infty$. Similarly, for the $\inf C = \inf {1/ x_1,0}=0$. Could you please verify? – darkmoor Aug 12 '17 at 09:11
  • @darkmoor Sorry, I was offline for 1-2 days. I will check your comments and write back to you – Jimmy R. Aug 14 '17 at 09:22
  • @darkmoor. See my edit. It seems that you are confused between $f(0)$ and $\lim_{x\to 0} f(x)$. The $\inf$ is attained exactly at $x=0$, (so it is actulally a $\min$, while the $\sup$ is attained as $x\to 0$ and it is indeed a $\sup$ (and not a $\max$) as it is attained asymptotically. – Jimmy R. Aug 14 '17 at 11:27
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    Thanks for your help and your time! – darkmoor Aug 16 '17 at 16:36
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Hint:

$1/x$ is a decreasing function.

Care have to be taken to handle $I_1$.

Siong Thye Goh
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For $i>1$, $f $ is stictly decreasing at $I_i=[x_{i-1},x_i] $ thus

$$M_i=f (x_{i-1})=\frac {1}{x_{i-1}} $$ $$m_i=\frac {1}{x_i} $$

but for $I_1$,

$$m_1=0$$ and $$M_1=\lim_{0^+}f (x)=+\infty $$

hamam_Abdallah
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