How to show that $f_n(x) = \frac{1}{1+nx^2}$ on $\mathbb{R}$ is not uniformly convergent

Question

I need to understand pointwise and uniform convergence of function series and have several exercise examples. Most of them I was able to solve but I got problems showing that:

$$ f_n(x) = \frac{1}{1+nx^2} $$

with $x \in \mathbb{R}$ is not uniformly convergent.

I will show you my thoughts and would be really happy if someone could point out my error.

I need to show that:

$$ |f_n(x)-f(x)|<\epsilon \text{ for } \forall\epsilon>0 $$

At first I calculate $f(x)$ as the limiting function, which should be:

$$ f(x) = \begin{cases} 1 & |x| < 1 \\ 0 & |x| \geq 1\end{cases} $$

Inserting the limiting function into the $\epsilon$ condition:

$$ \begin{split} |\frac{1}{1+nx^2}-1|&<\epsilon \text{ for } |x| < 1 \\ |\frac{1}{1+nx^2}-0|&<\epsilon \text{ for } |x| \geq 1 \end{split} $$ For the first case with $|x|<1$ the $\lim\limits_{n \rightarrow \infty}{\frac{1}{1+nx^2}}$ is $1$ and thus the inequation becomes $|1-1|<\epsilon$ for $\forall \epsilon>0$ which is true. For the second case with $|x| \geq 1$ the $\lim\limits_{n \rightarrow \infty}{\frac{1}{1+nx^2}}$ is 0 and thus the inequation becomes $|0-0|<\epsilon$ for $\forall\epsilon>0$ which is again true and thus the function series should be uniformly convergent. The solution to the exercise however tells me the function series is pointwise convergent but not uniformly convergent, so there must be some error in my thoughts.

Answer 1

The limit function is actually

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

Let $n \in \mathbb{N}$. Now, pick $x$ such that $|x| < \frac{1}{\sqrt{n}}$ and $x \neq 0$. Then,

$$x^2 < \frac{1}{n}$$ $$nx^2 < 1$$ $$1+nx^2 < 2$$ $$\frac{1}{1+nx^2}>\frac{1}{2}$$

Since the quantity on the left is positive, we can write

$$\left|\frac{1}{1+nx^2}\right|>\frac{1}{2}$$

That is,

$$\left|\frac{1}{1+nx^2}-0\right| > \frac{1}{2}$$

Which further implies that

$$|f_n(x)-f(x)| > \frac{1}{2}$$

So, it cannot be the case that $f_n$ converges uniformly to $f$.

Answer 2

The sequence of functions $(f_n)$ is point-wise convergent to the function $f$ defined by $$f(x)=0\;\;\text{if}\; x\ne0\quad\text{and}\;f(0)=1$$ and to prove that this convergent isn't uniform on $\Bbb R$ you have two ways:

• notice that the functions $f_n$ are continuous on $\Bbb R$ but $f$ isn't so we haven't the uniform convergence
• and the second way is $$\sup_{x\in\Bbb R}|f_n(x)-f(x)|\ge f_n\left(\frac1{\sqrt n}\right)=\frac12\xrightarrow{n\to\infty}\frac12\ne0$$ so the convergence isn't uniform on $\Bbb R$.