I am hoping that I could get feedback on my solution for the following problem.

Find the pointwise limit of the sequence $f_n$ and determine whether the convergence is uniform or not. $$f_n(x)=\begin{cases} nx^2 \quad & \text{if } \ x \in [0,\frac{1}{\sqrt{n}}], \\ 0 \quad & \text{if } \ x \in (\frac{1}{\sqrt{n}},1]. \\ \end{cases} $$

Note first that $f_n(0)=f_n(1)=0$. Let $x>0$, then $\exists N \in \mathbb{N}$ s.t $x>\frac{1}{\sqrt{N}}$ and $\forall n \geq N$, $x>\frac{1}{\sqrt{n}}$. Hence, for $0<x<1 \rightarrow f_n(x)=0$. Thus the pointwise limit of $f_n(x)$ is $f(x)=0$.

$\lim_{n \rightarrow \infty} \sup_{x \in [0,1]} |f_n(x)-f(x)|=1 \neq 0$. Therefore, $f_n(x) \nrightarrow^{u} f(x)$ on $[0,1]$.

Thank you for the feedback.

  • $\begingroup$ What is $f (1/\sqrt {n}) ?$ $\endgroup$ Jun 3, 2017 at 17:18
  • $\begingroup$ To me it's fine. You should precise which value takes $f_n$ at $\frac{1}{\sqrt{n}}$. $\endgroup$ Jun 3, 2017 at 17:20
  • $\begingroup$ @Joseph I ask the same question. what is $f (1/ sqrt n) $ ? $\endgroup$ Jun 3, 2017 at 17:32
  • $\begingroup$ @Joseph As it stands you have the functions taking two different values at $1/\sqrt{n}$ which means they aren't properly defined functions. Just cause it's discontinuous doesn't mean you can get away with it not being a well-defined function. (That being said, whether you handle the discontinuity by picking the right value or the left value doesn't particularly matter.) $\endgroup$ Jun 3, 2017 at 17:33
  • $\begingroup$ There was an error in the question, I have updated the function. It is defined to be 0 for $x \in (\frac{1}{\sqrt{n}},1]$. $\endgroup$
    – Joe
    Jun 3, 2017 at 17:41

1 Answer 1


hint for U convergence

for $n>0$,

$$f_n (\frac {1}{\sqrt {n}})=1$$ and $$f (\frac {1}{\sqrt {n}})=0$$

$$M_n=\sup_{x\in[0,1]}|f_n(x)-f (x)|\geq $$ $$|f_n (\frac {1}{\sqrt {n}})-f (\frac {1}{\sqrt{n}})|=1$$

$$\implies \lim_{n\to+\infty}M_n\ge 1$$ $$\implies \lim_{n\to+\infty}M_n \ne 0$$

the convergence is not uniform at $[0,1] $.

  • $\begingroup$ Thank you, your hint helps me with my understanding and gives me a different approach for similar problems, +1 and accepted answer. $\endgroup$
    – Joe
    Jun 3, 2017 at 19:23

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