So we had the pointwise and uniform convergence, and I do get that a sequence of function can converge to a function, just like ordinary sequences do. But what I don't quite get is this pointwise and uniform convergence...

For example, if we take the functions $$f_n,g_n,h_n:\mathbb{R}\rightarrow\mathbb{R} \qquad f_n(x)=\frac{x^{2n}}{1+x^{2n}}, g_n(x)=e^{-n\cdot x^2},h(x)=\sqrt{\frac{1}{n}+x^2}.$$ I have calculated the functions to which they converge pointwise: $$f(x) = \begin{cases} 0 & \textrm{ for $\lvert x\rvert<1$}\\ \frac{1}{2} & \textrm{ for $\lvert x\rvert=1$} \\ 1 & \textrm{ for $\lvert x\rvert>1$}\\ \end{cases}, g(x)=0,h(x)=\lvert x \rvert.$$

The only thing left is to see if they also converge uniform.

For $g_n(x)$ it's not hard to see that for any $\epsilon<\frac{1}{e}$ we have that $$\lvert f_n(\frac{1}{\sqrt{n}})-f(\frac{1}{\sqrt{n}})\rvert=e^{-1}>\epsilon,$$ which should be enough to show that it doesnt converges uniform.

For the other two functions I don't know how to prove or disprove their uniform convergence, but I do have a feeling that they are both uniformly convergent...

Is there a trick that I can use to determine here for determing the uniform convergence? Is there any relation to the norm $\lvert\lvert a_n(x)-a(x)\rvert\rvert$, where $a_n(x)$ converges to $a(x)$ pointwise?

  • $\begingroup$ Note $g_n(0)=1$ for every $n$, so the pointwise limit is not continuous. $\endgroup$
    – grand_chat
    May 2, 2016 at 23:25

1 Answer 1


$f_n$ converges to $f$ uniformly if and only if the supremum norm $\| f_n - f \|$ goes to zero. This isn't that useful in problems, though, because the supremum norm is usually hard to precisely calculate. You usually want to get some estimate of it rather than exactly computing it.

One theorem that you may not have proven yet is that the uniform limit of a sequence of continuous functions is continuous. That means that if the pointwise limit of a sequence of continuous functions is not continuous, then the convergence isn't uniform. Even without that theorem at your disposal in proofs, you can use it to determine what the answer is, which is always helpful in starting a "prove or disprove" problem.

In particular, in your $g_n$ problem, this suggests that you should look at points which get closer and closer to $0$ as $n \to \infty$, as you did. In your $f_n$ problem it similarly suggests that you look at points which get closer and closer to $1$ as $n \to \infty$.

In your $h_n$ problem, you have probably already seen a trick in calculus which is useful here: $\sqrt{1/n+x^2}-\sqrt{x^2}=\frac{1/n+x^2-x^2}{\sqrt{1/n+x^2}+\sqrt{x^2}}$. What does that give you in this case?

  • $\begingroup$ Well, I calculated that $\| h_n-h \|=\frac{1}{\sqrt{n}}$ and so it should converge uniformly bcs the limit of it is 0. Similar I also calculated that $\| f_n-f \|=\frac{1}{2}$ when $|x|\neq 1$ so it shouldnt converge uniformly... $\endgroup$
    – CryoDrakon
    May 2, 2016 at 23:24
  • $\begingroup$ @HeatTheIce Looks reasonable. $\endgroup$
    – Ian
    May 2, 2016 at 23:32

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