Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f_n : (a,b) \to \mathbb{R}$ be functions that have finite number of maxima and minima, for $n = 1,2,3...$. Let D be a countable dense subset of $(a,b)$. If sequence $\{f_n\}$ converges to $f$ such that the convergence is non-uniform at the points $x \in D$ and uniform at the points $x \in (a,b)\setminus D$, then does it imply that $f(x) = 0 \forall x \in (a,b)\setminus D$ ?

EDIT : The question didn't come out as what i expected. So I am posting a new one. sorry for the inconvinience.

share|cite|improve this question
"uniform at the points $x \in (a,b)\setminus D$" seems to be a bit of an unfortunate formulation. Do you mean to say "uniformly on $(a,b) \setminus D$"? – t.b. May 13 '11 at 8:25
@Theo : there seems to be a problem – Rajesh Dachiraju May 13 '11 at 8:27
@Rajesh: Does non-uniform mean pointwise convergence – user9413 May 13 '11 at 8:28
Seems I posted my comment as an answer. Anyway, @Theo, do you understand "non-uniform" here? Pointwise but not uniform? – Glen Wheeler May 13 '11 at 8:29
@Theo, @Glen : I will reformulate with your few minutes – Rajesh Dachiraju May 13 '11 at 8:29
up vote 1 down vote accepted

Certainly not. Perhaps you might want to further quantify what you mean by "non-uniform". Note that the sequence of functions $f_i(x) = c$ where $c$ is a fixed non-zero number converges pointwise (as well as uniformly, of course), and is a counterexample.

share|cite|improve this answer

Your Answer


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.