# Clarification on Pointwise and Uniform convergence

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.

## 1 Answer

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 $$\lim_{n\to\infty}\sup_{x\in I}|f_n(x)-f(x)|=0$$

• Thanks for the counterexample. This means my steps are not correct – Silver moon Feb 9 '15 at 5:07
• Yes, I understand. This is because integration will be equal if sequence of functions is uniformly convergent and not vice-versa – Silver moon Feb 9 '15 at 5:09
• Exactly, uniform convergence implies the convergence of integral, but not the other way(even with the point wise convergence) – Frank Lu Feb 9 '15 at 5:09
• 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/… – Silver moon Feb 9 '15 at 5:12
• 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 – Frank Lu Feb 9 '15 at 5:14