# Find the limit function of the given sequence $f_n$.

For each $$n \in \Bbb N$$, let $$f_{n}(x) = \begin{cases}nx, &0\leq x\leq \frac{1}{n}\\ 1, &\frac{1}{n} Find the limit function of the sequence $$f_{n}$$. Show that the convergence of the sequence is not uniform.

I am having difficulty in writing a formal proof.

As $$n$$ tends to $$\infty$$, we have $$x=0$$, so $$f(x)= 0$$. So limit function is $$f(x)= 0$$.

Next step is to show that the convergence is not uniform. Since for $$x=1$$ the condition $$|f_{n}(x) - f(x)|< \epsilon$$ is not satisfied if we take $$\epsilon$$ less than $$1$$, so the convergence is not uniform.

Is this approach correct?

Please guide me to write a formal proof for the first part.

### limit function $$f$$

For $$x=0$$ holds $$x \le \frac1n$$ for all $$n\in\mathbb N$$. So $$f_n(0) = 0$$ for all $$n\in\mathbb N$$. Thus $$f(0)=0$$.

For every $$x>0$$ exists a $$N\in\mathbb N$$ such that $$x>\frac1n$$ for all $$n>N$$ (Archimedean property). That is $$f_n(x) = 1$$ for all $$n>N$$. Thus $$f(x)=1$$.

So $$f(x) = \begin{cases}0 & x=0\\1 & 0

### no uniform convergence

Proof by contradiction: let be $$0<ε<\frac12$$. Assume there is a $$N\in\mathbb N$$ such that for all $$x\in[0,1]$$ and all $$n\ge N$$: $$\vert f_n(x)-f(x)\vert < ε.$$ Choose $$x=\frac{1}{2N}$$. Then $$f_N(x) = \frac12$$. So $$\vert f_N(x)-f(x)\vert = \vert \frac12 - 1\vert = \frac12 \not< ε$$ what is a contradiction.

• pH 74. Well done. Mar 5, 2020 at 9:50
• Thanks @PeterSzilas! I did not do such proofs for about 10 years. So it was a nice refreshment for myself… Mar 5, 2020 at 10:25
• pH74. I struggle a bit too:) Mar 5, 2020 at 10:48

Option.

For $$x >0$$:

Pick $$n_0$$ (Archimedean principle) s.t for $$n \ge n_0$$: $$1/n , then $$f_n(x)=1$$.i.e. $$\lim_{n \rightarrow \infty}f_n(x)=1$$ for $$x>0$$.

For $$x=0$$: $$\lim_{n \rightarrow \infty}f_n(0)=0.$$.

$$f_n$$ are continuos on $$[0,1]$$.

If $$f_n$$ were uniformly convergent the limit function $$f$$ would be continuos. Hence?

• $f$ is discontinuous, so convergence must be non-uniform. Mar 5, 2020 at 9:15
• Mathsaddict.Yes, if f not continuos, convergence is not uniform. Mar 5, 2020 at 9:21

$$\lim f_n(0)=0$$ is clear. For $$x>0$$, $$|f_n(x)-1|=0<\epsilon$$ if $$n >\frac 1 x$$. So $$f_n(x) \to f(x)$$ for every $$x$$ where $$f(0)=0$$ and $$f(x) =1$$ for $$x >0$$.

If $$0<\epsilon <1/2$$ then $$|f_n(\frac 1 {n^{2}}) -f(\frac 1 {n^{2}})|>\epsilon$$ for every $$n\geq 2$$. Hence the convergence is not uniform.

• Kavi.Second last line $f(1/n^2)$. Mar 5, 2020 at 9:24
• @PeterSzilas Yes, thank you for pointing out. Mar 5, 2020 at 9:30