Question related to Riemann sums, sups, and infs of bounded functions

  • #1
Can someone give me an example of a bounded function [itex]f[/itex] defined on a closed interval [itex][a,b][/itex] such that [itex]f[/itex] does not attain its sup (or inf) on this interval? Obviously, [itex]f[/itex] cannot be continuous, but for whatever reason (stupidity? lack of imagination?) I can't think of an example of a discontinuous, bounded function for which [itex]\sup\limits_{x\in [a,b]} f(x) \neq \max\limits_{x\in [a,b]} f(x)[/itex].
 

Answers and Replies

  • #2
lavinia
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Can someone give me an example of a bounded function [itex]f[/itex] defined on a closed interval [itex][a,b][/itex] such that [itex]f[/itex] does not attain its sup (or inf) on this interval? Obviously, [itex]f[/itex] cannot be continuous, but for whatever reason (stupidity? lack of imagination?) I can't think of an example of a discontinuous, bounded function for which [itex]\sup\limits_{x\in [a,b]} f(x) \neq \max\limits_{x\in [a,b]} f(x)[/itex].

f(x) = x except at the end points where it equals (a+b)/2
 
  • #3
Hmmm...I think I may have just thought of one! On [itex][0,1][/itex], define [itex]f(1) = 0[/itex], and let
[tex]
f(x) = \sum_{n = 0}^\infty \frac{n}{n+1} \chi_{\left[ \left. \frac{n}{n+1},\frac{n+1}{n+2} \right) \right.}(x).
[/tex]
Then [itex]f(x) \leq 1[/itex] for all [itex]x[/itex], [itex]\sup\limits_{x\in[0,1]} f(x) = 1[/itex], but [itex]f(x) \neq 1[/itex] for all [itex]x\in [0,1][/itex]. And, if I'm not mistaken, [itex]f[/itex] is still Riemann integrable, since there are only countably many points at which it is discontinuous. Can someone confirm if this is actually true?
 
  • #4
f(x) = x except at the end points where it equals (a+b)/2
This is much easier than my example. Thanks!
 

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