Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $I\subseteq\mathbb{R}$ be a closed and bounded interval and let $$f:I\to(0,\infty)$$ be a bounded function that is discontinuous at every point in $I$.

Does there exist a bounded function $g:I\to(0,\infty)$ such that the product $fg$ is continuous at least at one point in $I$?

share|improve this question
3  
How about $g=17/f$? –  Gerry Myerson Feb 28 '13 at 0:23
    
Oups, I forgot to specify bounded... Let me edit. –  Spenser Feb 28 '13 at 0:25
    
What have you tried? –  Trevor Wilson Feb 28 '13 at 0:27

1 Answer 1

up vote 2 down vote accepted

Not necessarily. Let $f(x) = 1/2^n$ if $x = m/2^n$ for some $m \in \mathbb{N}$, and 1 otherwise. If $fg$ is continuous at $x_0$, then for some $a>0$ we have $g(x) \approx a/f(x)$ near $x_0$. Then you can show that $g$ is unbounded. (In fact, it is unbounded in every neighborhood of $x_0$.)

Do you see how to make this argument precise?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.