# Thread: Lower integral on two functions

1. ## Lower integral on two functions

Suppose that two bounded functions $\displaystyle f : [a,b] \rightarrow R$ and $\displaystyle g : [a,b] \rightarrow R$ have the property that $\displaystyle g(x) \leq f(x)$ for all x in [a,b].

For a partition P of [a,b] show that $\displaystyle L(g,P) \leq L(f,P)$

I tried to do this by considering a partition $\displaystyle P = \{{x_0},...{x_n}\}$, and for each index $\displaystyle i \geq 1$, the number m is a lower bound for both the functions given x in $\displaystyle [{x_i-1}, {x_i}]$. so, $\displaystyle m \leq m_i$
is this the correct path? I am stuck on this step..

2. Originally Posted by serious331
Suppose that two bounded functions $\displaystyle f : [a,b] \rightarrow R$ and $\displaystyle g : [a,b] \rightarrow R$ have the property that $\displaystyle g(x) \leq f(x)$ for all x in [a,b].

For a partition P of [a,b] show that $\displaystyle L(g,P) \leq L(f,P)$

I tried to do this by considering a partition $\displaystyle P = \{{x_0},...{x_n}\}$, and for each index $\displaystyle i \geq 1$, the number m is a lower bound for both the functions given x in $\displaystyle [{x_i-1}, {x_i}]$. so, $\displaystyle m \leq m_i$
is this the correct path? I am stuck on this step..
What are you having trouble with, precisely?

3. Originally Posted by Drexel28
What are you having trouble with, precisely?
What I am unsure about is that if we are considering the same partition, as defined above, for both the functions, then for both the functions we end up as:

$\displaystyle m(b-a) = \sum_{i=1}^n m({x_i} - {x_{i-1}}) \leq \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(g,P)$

and $\displaystyle \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(f,P)$

Is that what I need to do?

4. Originally Posted by serious331
What I am unsure about is that if we are considering the same partition, as defined above, for both the functions, then for both the functions we end up as:

$\displaystyle m(b-a) = \sum_{i=1}^n m({x_i} - {x_{i-1}}) \leq \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(g,P)$

and $\displaystyle \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(f,P)$

Is that what I need to do?
It seems like you're all messed up.

Would you agree that $\displaystyle f\leqslant g\implies \inf_{x\in I}f(x)\leqslant\inf_{x\in I}g(x)$ for any $\displaystyle I\subseteq [a,b]$?

5. Originally Posted by Drexel28
It seems like you're all messed up.

Would you agree that $\displaystyle f\leqslant g\implies \inf_{x\in I}f(x)\leqslant\inf_{x\in I}g(x)$ for any $\displaystyle I\subseteq [a,b]$?

I think I would agree to the inequality you have presented, but I dont think I wont be able to write a proof on how that is true!

isnt it $\displaystyle {m_{i_g}} \leq {m_{i_f}}$?

6. If $\displaystyle g\le f$, you have that for any $\displaystyle x\in I$, $\displaystyle \inf_{y\in I} g(y)\le g(x) \le f(x)$, so $\displaystyle \inf_{y\in I} g(y)$ is a lower bound for $\displaystyle f(y)$ ($\displaystyle y\in I$). Since $\displaystyle \inf_{y\in I} f(y)$ is the greatest lower bound of the f(y), it follows that it is $\displaystyle \ge$ the lower bound $\displaystyle \inf_{y\in I} g(y)$.

7. Originally Posted by serious331
Suppose that two bounded functions $\displaystyle f : [a,b] \rightarrow R$ and $\displaystyle g : [a,b] \rightarrow R$ have the property that $\displaystyle g(x) \leq f(x)$ for all x in [a,b].

For a partition P of [a,b] show that $\displaystyle L(g,P) \leq L(f,P)$

I tried to do this by considering a partition $\displaystyle P = \{{x_0},...{x_n}\}$, and for each index $\displaystyle i \geq 1$, the number m is a lower bound for both the functions given x in $\displaystyle [{x_i-1}, {x_i}]$. so, $\displaystyle m \leq m_i$
is this the correct path? I am stuck on this step..
For any partition interval $\displaystyle [x_{i-1}, {x_i}]$:
$\displaystyle m_{g_i} = \mbox{inf} \{g(x) : x \in [{x_{i-1}},{x_i}]\} \leq m_{f_i} = inf \{f(x) : x \in [{x_{i-1}},{x_i}]\}$
So, for each $\displaystyle x \in [{x_{i-1}},{x_i}]$, we have $\displaystyle g(x) \leq f(x)$. Suppose this is false & $\displaystyle {m_{g_i}} > {m_{f_i}}$
Let $\displaystyle \epsilon = \frac{{m_{g_i}} - {m_{f_i}}}{2}$. Then, there exists $\displaystyle x \in [{x_{i-1}},{x_i}]$ and f(x) such that $\displaystyle f(x) - {m_{f_i}} < \epsilon \implies f(x) < \epsilon + {m_{f_i}} = {m_{f_i}}+ \frac{{m_{g_i}} - {m_{f_i}}}{2} = \frac{{m_{g_i}} + {m_{f_i}}}{2} < {m_{g_i}}$ which is a contradiction!