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 $I$?

  • 3
    $\begingroup$ How about $g=17/f$? $\endgroup$ – Gerry Myerson Feb 28 '13 at 0:23
  • $\begingroup$ Oups, I forgot to specify bounded... Let me edit. $\endgroup$ – Spenser Feb 28 '13 at 0:25

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?


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