# Prove that the sequence does not converge uniformly

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 $$0 $$\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!

• 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? Sep 24 at 21:23
• 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. Sep 24 at 21:25
• 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. Sep 24 at 21:25
• The conclusion about the integral can be obtain more naively than by these two methods:$$0\le\int_0^1f_n(x)dx\le\int_0^{1/n}dx+\frac1{n^2}\int_{1/n}^1\frac{dx}{x^2}=\frac1n+\frac1{n^2}(n-1).$$ Sep 24 at 21:32

• 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: $$0\le\int_0^1f_n(x)dx\le\int_0^{1/n}dx+\frac1{n^2}\int_{1/n}^1\frac{dx}{x^2}=\frac1n+\frac1{n^2}(n-1).$$

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

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

Then

\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