0
$\begingroup$

Let $g:[a.b] \rightarrow \mathbb{R}$ be a bounded function. Prove that $U(g)$ and $L(g)$ exist and $L(g) \leq U(g)$.

I am new to this material and am confused on how to prove this statement. We know:

$L(g) = \sup\{L(g , P) :$ P is a partition of $[a,b]\}$.

$U(g) = \inf\{U(g , P) :$ P is a partition of $[a,b]\}$.

Note that \begin{align*} U(g , P) = \sum_{i=1}^{n} M_i(x_i - x_{i-1}) \end{align*} and \begin{align*} L(g , P) = \sum_{i=1}^{n} m_i(x_i - x_{i-1}) \end{align*} where $M_i = \sup\{g(x) : x \in [x_{i-1}, x_i]\}$ and $m_i = \inf\{g(x) : x \in [x_{i-1}, x_i]\}$.

Any hints will be appreciated.

  • 0
    While many of us know what you mean by $L(g,P)$ and $U(g,P)$, you should define them regardless (a descriptive sentence will do). – 2017-01-22
  • 1
    Hint: $M(b-a)$ is an upper bound for $L(g)$, where $|f|$x$, and every non-empty set in $\mathbb{R}$ that has an upper bound has a sup. – 2017-01-22

3 Answers 3

0

Since $g$ is bounded, say $|g(x)| \le B$, then $L(g,P) \le B(b-a)$ for any partition $P$, hence $L(g) \le B(b-a)$. Similarly, $U(g) \ge -B(b-a)$.

It is straightforward to see that $L(g,P) \le U(g,P)$ for any given partition $P$, and if $P'$ is a refinement of $P$, then $L(g,P) \le L(g,P') \le U(g,P') \le U(g,P)$.

In particular, if $P_1,P_2$ are partitions, let $P''$ be refinement of both, then $L(g,P_1) \le L(g,P'') \le U(g,P'') \le U(g,P_2)$, in particular, $L(g,P_1) \le U(g,P_2)$ for any two partitions $P_1,P_2$.

Hence it follows that $L(g) \le U(g,P_2)$ for any partition $P_2$ and hence $L(g) \le U(g)$.

0

Hint: First, show that $$|L(g,P)|,|U(g,P)| \leq (b-a)\sup|g| \qquad \text{for all partitions } P.$$ Use this to conclude that $L(g)$ and $U(g)$ are real numbers. Now, show that $$L(g,P) \leq U(g,P) \qquad \text{for all partitions } P.$$ Use this to obtain $L(g) \leq U(g)$.

0

Observe that $-(b-a)M \le L(g,P)$, and $U(g,P) \le <(b-a)M$ for all $P's$, thus the $L(g,P)$'s and the $U(g,P)$'s are bounded below and above by $-(b-a)M$, and $(b-a)M$ with $M$ being $|f(x)| \le M$ as $f$ is bounded. Thus there exists a $inf$ and $sup$ for them respectively. The $L(g,P) \le U(g,P)$ follows from the fact that $\inf_{g}\le \sup_{g}$ on any partition interval $[x_{i-1}, x_i]$