I want to prove by definition that the following sequence of functions

$${f}_{n}= {1 \over 1+n^2x^2}$$

$x\in [0, 1]$, $n\in \mathbb{N}$, converges uniformly in the interval $(0, 1]$

We can find the pointwise limit $\lim \limits _{n\to \infty} f_n(x_{o})=\lim \limits _{n\to \infty} {1 \over 1+n^2x_{0}^2}$

Consider the following cases

  1. If $x_{0}=1$ $\rightarrow$ $\lim \limits _{n\to \infty} f_n(x_{o})=\lim \limits _{n\to \infty} {1 \over 1+n^2x_{0}^2}=0$

  2. If $0<x_{0}<1$ $\rightarrow$ $\lim \limits _{n\to \infty} f_n(x_{o})=\lim \limits _{n\to \infty} {1 \over 1+n^2x_{0}^2}=0$

We can see that the pointwise limit is $\lim \limits _{n\to \infty} f_n(x)=f(x)=0$ for $x\in (0, 1]$

Now, seeing that the convergence is not uniform has been difficult for me.

My idea is that as the space of bounded real functions $(\mathbf{B}_{\mathbb{R}}((0, 1]), ||.||_{\infty})$, from where $||f||_{\infty}=\sup_{x \in (0, 1]} \{|f_{n}(x)| \}$, is complete then the limit function $f(x)$ should belong to $\mathbf{B}_{\mathbb{R}}((0, 1])$ and reach some kind of contradiction, but I don't see it. Any suggestions?

In addition, I also want to show that

$$\lim \limits _{n\to \infty} \int_{0} ^{1}f_n(x)=\int_{0} ^{1} \lim \limits _{n\to \infty} f_n(x)$$

But taking into account that the pointwise limit is $0$

It all comes down to showing that

$$\lim \limits _{n\to \infty} \int_{0} ^{1}f_n(x)=0$$

Note that

$$\lim \limits _{n\to \infty} \int_{0} ^{1}f_n(x) \, dx=\lim \limits _{n\to \infty} \int_{0} ^{1}{1 \over 1+n^2x^2} \, dx=\lim \limits _{n\to \infty} \frac{1}{n} \arctan(nx)\Bigg|_{0}^{1}$$

So $$\lim \limits _{n\to \infty} \int_{0} ^{1}f_n(x)=\arctan \cdot \lim \limits _{n\to \infty} \frac{1}{n}=0 $$

Is this conclusion correct?

Any help would be appreciated!

  • $\begingroup$ Your first sentence contradicts your title. $\endgroup$ Sep 24 at 21:21
  • 1
    $\begingroup$ I'm not sure what you're getting at because you've shown what the limit function should be: it's obviously bounded. In fact, all of the $f_n$ are uniformly bounded and take values in $[0,1]$. Think about if our domain was $[0,1]$. Where does uniform convergence go wrong, and how do we show that? $\endgroup$
    – 1mdlrjcmed
    Sep 24 at 21:23
  • $\begingroup$ If the convergence was uniform on $(0,1],$ the sequence $(f_n)$ would be uniformly Cauchy on $(0,1]$ hence (by continuity) also on $[0,1].$ But the pointwise limit on $[0,1]$ is not continuous at $0.$ Contradiction. $\endgroup$ Sep 24 at 21:25
  • 1
    $\begingroup$ The conclusion about the integral follows from the dominated convergence theorem, but what you did also works provided you fix the typo on the last line. $\endgroup$
    – 1mdlrjcmed
    Sep 24 at 21:25
  • $\begingroup$ The conclusion about the integral can be obtain more naively than by these two methods:$$0\le\int_0^1f_n(x)dx\le\int_0^{1/n}dx+\frac1{n^2}\int_{1/n}^1\frac{dx}{x^2}=\frac1n+\frac1{n^2}(n-1).$$ $\endgroup$ Sep 24 at 21:32

4 Answers 4

  • If the convergence was uniform on $(0,1],$ i.e. $$\forall\epsilon>0,\exists N,\forall n\ge N,\forall x\in(0,1],|f_n(x)|\le\varepsilon$$ the sequence $(f_n)$ would be uniformly Cauchy on $(0,1]$ i.e. $$\forall\eta>0,\exists N,\forall p,q\ge N,\forall x\in(0,1],|f_p(x)-f_q(x)|\le\eta$$ hence (by continuity of these functions on $[0,1]$) $$\forall\eta>0,\exists N,\forall p,q\ge N,\forall x\in[0,1],|f_p(x)-f_q(x)|\le\eta$$ i.e. on $[0,1],$ $(f_n)$ would be uniformly Cauchy hence uniformly convergent. But it is not, since the $f_n$'s are continuous, whereas the pointwise limit on $[0,1]$ is not continuous at $0.$

  • A more specific and direct argument is: the uniform convergence on $(0,1]$ would entail $$\exists n\in\Bbb N,\forall x\in(0,1],\frac1{1+n^2x^2}\le\frac12$$ i.e. $$\exists n\in\Bbb N,\forall x\in(0,1],x\ge\frac1n,$$ which is clearly false.

  • As for your conclusion about the integral, it is correct (up to the typo in the last line) but can be obtained more naively: $$0\le\int_0^1f_n(x)dx\le\int_0^{1/n}dx+\frac1{n^2}\int_{1/n}^1\frac{dx}{x^2}=\frac1n+\frac1{n^2}(n-1).$$

  • $\begingroup$ This is surely a duplicate. $\endgroup$ Sep 24 at 23:15
  • $\begingroup$ @geetha290krm I had the same feeling but all the posts I found where about the (non) uniform convergence on $[0,1],$ not on $(0,1].$ $\endgroup$ Sep 25 at 5:50

Given $x_0 > 0$, and fixed $\epsilon > 0$, which is the smallest n such that $1 / (1 + n^2 \cdot x_0^2) < \epsilon$ ?

We need $n > \sqrt {1 / \epsilon - 1} / x_0$, so no fixed $n$ will succeed for arbitrarily small $\epsilon$.


Here is a slightly more direct argument that $f_n$ cannot converge uniformly on $(0,1]$:

If $f_n(x)$ converges uniformly on $(0,1]$ then it must converge to its pointwise limit, the zero function, and so there must exist an $N \in\mathbb N$ such that for all $n \geq N$ we $|f_n(x)|<1/3$. But for any $n\geq N$ we would then have $\lim_{x \to 0} |f_n(x)|\leq 1/3$ which is a contradiction, since the $\lim_{x \downarrow 0} f_n(x)=1$.


By negating the definition of uniform convergence, we find that for a sequence of functions $f_n : I\to\mathbb{R} $ to not converge uniformly to the pointwise limit $f$ there must be some $\varepsilon_0$ and a subsequence of functions

$$ f_{n_k} : I \to \mathbb{R} $$

And some sequence $x_k$ in $I$ such that

$$ | f_{n_k}(x_k) - f(x_k) | \geq \varepsilon_0 $$

For any $k$.

We can let $f_{n_k} = f_k $ (i.e. we take the entire sequence) and then define

$$ x_k = \frac{1}{k} $$


\begin{align} | f_{n_k}(x_k) - f(x_k) | &= \left|\frac{1}{1+\frac{k^2}{k^2}} - 0\right| \\\\ &= \frac{1}{2} \geq \frac{1}{2} \end{align}

So the convergence is non-uniform.

See chapter 8 of Bartle and Sherbert's introduction to real analysis For more information about the sequence criterion


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