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let $f(x)$ be an incresing function, and let $C$ be a constant s.t. $f(x) \leq C.$ Put $D = $ sup $f(x)$, and i need to show that $f(x) \rightarrow D$ as $x \rightarrow \infty$. It seems 'obvious' when i draw a picture, but hints for a formal proof would be nice. Thank you

Tom
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2 Answers2

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Hint: For any $\epsilon,$ we have that $D-\epsilon$ is not an upper bound of the range of $f,$ so there exists some $x_0$ such that $$D-\epsilon<f(x_0)\le D.$$ Can you take it from there?

Cameron Buie
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$\textbf{Hints:}$ Since $f$ is increasing and bounded from above, the limit $\lim_{x \to \infty} f(x)$ exists. What happens if the limit value is smaller or larger than $D$?

  • Well, if the limit value is larger than $D$, then $D$ is not an upper bound of $f(\mathbb{R})$.
  • However, if the limit value is smaller than $D$, then $D$ is not the $\textit{smallest}$ upper bound of $f(\mathbb{R})$.

Can you make this explicit?

user133281
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  • Good point, so the conclusion is that (given the two statements you wrote) that f(x) indeed tends to D, as x tends to \infty, else D would not be the supremum of f(x)? – Tom Mar 17 '14 at 17:02
  • Exactly. If $f(x)$ tends to a value larger than $D$, then in the range of $f$ there are values larger than $D$ (which is not possible if $D = \sup f$). If $f(x)$ tends to a value $L$ smaller than $D$, then for instance $\frac{L+D}{2}$ is an upper bound for $f(\mathbb{R})$ dat is smaller than $D$, contradicting the fact that $D$ is the supremum of $f$. – user133281 Mar 17 '14 at 17:22