$f_n:[0,1]\rightarrow \mathbb{R}$ and $f_n(x)=\begin{cases} n^2x & \text{ if } 0\leq x< \frac{1}{n} \\ \frac{1}{(n+1)x-1}& \text{ if } \frac{1}{n}\leq x\leq 1 \end{cases}$

Study convergence and the uniformly convergence.

for $x=0$ we have $f_n(0)=0$ for $0< x\leq 1$ it exists a $n_o \in\mathbb{N}$ such that $\frac{1}{n_0}<x \Rightarrow $ for every $n>n_0$ $\frac{1}{n}<x\Rightarrow $ $f_n(x)=\frac{1}{(n+1)x-1}\rightarrow 0$ as $n\rightarrow \infty $

$\Rightarrow f_n\rightarrow0$

I want to show that $f_n $ doesn't convergence uniformly, I calculated $\int_{0}^{1}f_ndx=1/2\neq 0 $ and that mean that indeed it doesn't convergence uniformly. But it $\int_{0}^{1}f_ndx=0$ does that mean that $f_n $ convergence uniformly? Are there any other ways to show that it is uniformly converges or not. how can I use the $sup$ on a piecewise $f_n $?


2 Answers 2


I think that it is much simpler to say that$$\sup_{x\in[0,1]}|f_n|=n.\tag1$$So, you don't have $\lim_{n\to\infty}|f_n-0|=0$, which means that the convergence is not uniform.

Note that $f_n\left(\frac1n\right)=n$, and this is enough to prove that $\sup_{x\in[0,1]}|f_n|\geqslant n$, which, in turn, is enough to prove that the convergence is not uniform. You actually have $(1)$ because $f$ is increasing on $\left[0,\frac1n\right]$ and decreasing on $\left[\frac1n,1\right]$.

On the other hand, even if $(f_n)_{n\in\Bbb N}$ is a sequence of continuous functions from $[0,1]$ into $\Bbb R_+$ which converges pointwise to the null function and that $\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm dx=0$, you cannot deduce that the convergence is uniform. Take, for instance$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}nx&\text{ if }x\leqslant\frac1{2n}\\1-nx&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ otherwise.}\end{cases}\end{array}$$The convergence is not uniform because$$(\forall n\in\Bbb N):\sup f_n=f_n\left(\frac1{2n}\right)=\frac12.$$


For all $n$ you have $$f_n(1/n) = n$$ Hence $$\sup_{x \in [0,1]} |f_n(x)| \ge n$$ This means that $f_n$ does not converge uniformly.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .