If a sequence of functions $f_{n}$ converges pointwise to some function$f$, it doesn't imply uniform convergence but the converse is true.

My question is if we want to test the uniform convergence of a sequence of functions $f_{n}$ can we adopt the following proecdure:

Step1: Figure out the pointwise convergence of the sequence of functions, say $f_{n}$ converges to $f$.

Step2: Then find the integral of $f_{n}$ as well as integral of $f$. If the integrals are equal, then $f_{n}$ converges uniformly to $f$.

Is this a legit test? I am sure this is correct but I just want to clarify.


Consider $f_n:(-1,1)\to\mathbb{R}$, $f_n(x)=x^n$, then $f_n$ converge pointwise to $f(x)\equiv 0$. Moreover we have $$\lim_{n\to\infty}\int_{-1}^1f_n(x)dx=0$$

However, $f_n$ does not converge uniformly, the proof can be seen here

To determine whether the convergence is uniform, usually it is easy with the following criterion: suppose $f_n:I\to\mathbb{R}$ is a sequence of functions which converges point wise to a function $f$, then the convergence is uniform if and only if $$\lim_{n\to\infty}\sup_{x\in I}|f_n(x)-f(x)|=0$$

  • $\begingroup$ Thanks for the counterexample. This means my steps are not correct $\endgroup$ – Silver moon Feb 9 '15 at 5:07
  • $\begingroup$ Yes, I understand. This is because integration will be equal if sequence of functions is uniformly convergent and not vice-versa $\endgroup$ – Silver moon Feb 9 '15 at 5:09
  • $\begingroup$ Exactly, uniform convergence implies the convergence of integral, but not the other way(even with the point wise convergence) $\endgroup$ – Frank Lu Feb 9 '15 at 5:09
  • $\begingroup$ We both pointed to the same thing. I got confused because of this, please take a look at the first answer to this question math.stackexchange.com/questions/615399/… $\endgroup$ – Silver moon Feb 9 '15 at 5:12
  • $\begingroup$ This is like a counter positive of the result, because if the integral does not converge(to the integral of $f$) then the convergence cannot be uniform $\endgroup$ – Frank Lu Feb 9 '15 at 5:14

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.