Is Every Function with Equal Lower and Upper Sums Constant?

[evinda]

Hello!

Let $f:[a,b] \to \mathbb{R}$ function with the identity:
$$L(f,P)=U(f,P), \text{ for any partition } P \text{ of } [a,b]$$

Show that $f$ is a constant.

Can I do it like that?

We pick for the interval $[a,b]$ a partition of $2$ points: $P=\{a=x_0,x_1=b\}$

Then,it is like that: $L(f,P)=(b-a)inf([a,b]) \text{ and } U(f,P)=(b-a)supf([a,b])$

Since we know that $L(f,P)=U(f,P) \Rightarrow inf([a,b])=supf([a,b])=c \Rightarrow f(x)=c$

 

[Opalg]

Yes, that looks good. (Yes)

 

[evinda]

Yes, that looks good. (Yes)

Nice..Thank you! :)

 

[evinda]

Yes, that looks good. (Yes)

Is this partition the only possible that I could take? (Thinking)

 

[blue_raver22]

for all $x \in [a,b]$

Yes, you can approach it like that. Another way to prove that $f$ is a constant is by contradiction. Assume that $f$ is not a constant, then there exist $x,y \in [a,b]$ such that $f(x) \neq f(y)$. Without loss of generality, let's say $f(x) < f(y)$. Then, we can choose a partition $P$ such that $x \in [x_0,x_1]$ and $y \in [x_1,x_2]$, where $x_0,x_1,x_2$ are points in the partition. Since $f$ is continuous, it must attain its maximum and minimum values on $[x_0,x_1]$ and $[x_1,x_2]$ respectively. But this contradicts the fact that $L(f,P)=U(f,P)$, since $L(f,P)$ will be strictly less than $U(f,P)$. Therefore, our assumption that $f$ is not a constant must be false, and thus $f$ is a constant.