Suppose we have a bounded continuous function $f(x)$ on some interval $(a,b)$. Suppose we also have an function $g(x)$ that is uniformly continuous on the same interval $(a,b)$. Then, is the product $f(x)g(x)$ uniformly continuous? Intuition tells me that it is true. However, I am not sure. Most of the cases that I've dealt with involve relating an explicit formula for the function (the domain) to an epsilon and a delta. I have tried using the same approach as done in product of two uniformly continuous functions is uniformly continuous

However, I am stuck at the same part. I do not know how to bound them.

  • 9
    $\begingroup$ You may start with $g\equiv 1$. $\endgroup$ – 23rd Oct 18 '13 at 19:39
  • 2
    $\begingroup$ An English usage issue arises. When a word that starts with a sound similar to that of "you", such as "uniform", "union", "Europe", etc., is preceded by an indefinite article, then the form it takes is "a" rather than "an". (One can find books published in the 18th century that say "an union", etc.) I changed the title accordingly. $\endgroup$ – Michael Hardy Oct 18 '13 at 19:48
  • 1
    $\begingroup$ What Landscape is slyly suggesting is that the statement is false. $\endgroup$ – user43208 Oct 18 '13 at 19:51
  • $\begingroup$ Yes, I know. I think he should have made it an answer. $\endgroup$ – CoffeeIsLife Oct 18 '13 at 19:54
  • $\begingroup$ Not necessarily. The OP, on the other hand, should not let this question hanging: either close it or accept the current answer. $\endgroup$ – Did Aug 18 '14 at 9:07

This is not true.

Let $g(x)=1$, the interval $I$ be $(0,1)$ and $f(x)=\sin\left(\dfrac{1}{x}\right)$.

Then applying the theorem would imply that $f(x)=\sin\left(\dfrac{1}{x}\right)$ is uniformly continuous on $(0,1)$ which is incorrect(See here for proof).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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