The Question: Determine whether the sequence converges uniformly:

${f_n(x)}: [0,1] \rightarrow \mathbb{R}$,

$f_n(x) = \begin{cases} 0 & f_n(0) = f(2/n) = f(1) = 0 \\ n & x = 1/n \\ \end{cases} $ , and $f$ is linear on the intervals $[0,1/n]$, $[1/n,2/n]$, and $[2/n, 0]$.

The Attempt: Here is a graph of the function: enter image description here

There is a typo in the picture. The coordinate $(1/n, 1)$ should be $(1/n,n)$. It was proven that the sequence of functions converges point wise to $f(x) = 0$. It seems obvious to me that the sequence is not uniformly convergent.

Let $N \geq 1$. Choose $\epsilon_0 = 1/3$, Then let $n_0 = N$, and $x= 1/n_0$. Then, $|f_n(x) - f(x)| = |f_n(1/n_0) - f(1/n_0)| = |n_0 - 0| = n0 \geq N > 1/3$. Hence the sequence of functions is not uniformly convergent?

Is this correct or do I need some detail?

Thank you very much!!

  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$
    – Clement C.
    Jan 26, 2017 at 23:00

1 Answer 1


What you are doing with $n_0$ and $N$ is a bit confusing: I assume you are trying to negate the definition of convergence to show that the sequence $(s_n)_n$ defined by $$ s_n\stackrel{\rm def}{=} \sup_{x\in[0,1]} \lvert f_n(x)-f(x)\rvert$$ does not converge to $0$ (and thus that $(f_n)_n$ does not converge uniformly to $f\equiv0$).

In that case, the gist of what you are doing is correct, but it would be, in my opinion, must clear to write the following: let $x_n\stackrel{\rm def}{=} \frac{1}{n} \in[0,1]$ for $n\geq 1$. By definition of the supremum, for any $n\geq 1$ we have $$ s_n \geq \lvert f_n(x_n)-f(x_n)\rvert = \lvert f_n(x_n)\rvert = n $$ and therefore $s_n\xrightarrow[n\to\infty]{} \infty$ (and clearly not $0$).

Now, as a followup question: does it converge uniformly on $(0,1]$? And, more interestingly, on $[a,1]$ for any fixed $a>0$?

  • $\begingroup$ Well I am trying to use the negation of the $\epsilon-N$ definition of convergence. I really don't understand the supremum definition of uniform convergence. What does it mean exactly? $\endgroup$
    – user211962
    Jan 22, 2017 at 21:00
  • $\begingroup$ The first point I have to check: this is how uniform convergence has been introduced to you, right? (by its definition) $\endgroup$
    – Clement C.
    Jan 22, 2017 at 21:01
  • $\begingroup$ Basically, this means that if you were to draw/plot the limiting function $f$ on $[0,1]$, as well as those of the of the $f_n$'s (infinitely many of them), then the graphs of the $f_n$'s converge to/approach the graph of $f$: the maximum distance between the graph of $f_n$ and the graph of $f$ over all points $x\in[0,1]$ tends to $0$. $\endgroup$
    – Clement C.
    Jan 22, 2017 at 21:04
  • $\begingroup$ Well I try think about it is if the sequence of functions are well inside an epsilon tube for all points in the domain, then the function is uniformly convergent. For a sequence of functions not to be uniformly convergent implies for some points in the domain, they are outside of the epsilon tube. I am trying to apply this in my proof. $\endgroup$
    – user211962
    Jan 22, 2017 at 21:13
  • $\begingroup$ The minimum width of the "tube" necessary for $f_n$ to be in the tube around $f$ is what I called $s_n$. You want this width to converge to $0$ (or, with the equivalent $\varepsilon$-formulation of your comment, to be less than $\varepsilon$ for all $n\geq N_\varepsilon$, where $\varepsilon > 0$ is arbitrary and $N_\varepsilon$ depends on it). $\endgroup$
    – Clement C.
    Jan 22, 2017 at 21:16

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