# Construct sequence of continuous functions such that $\lim_{n \to \infty} \int _0 ^ 1 f_n(x) \,dx = +\infty$

Construct sequence of continuous functions $f_n:[0,1] \to \mathbb{R}$ such that $\displaystyle \lim _{n \to \infty} f_n(x)=0$ implies that $\displaystyle \lim_{n \to \infty} \int _0 ^ 1 f_n(x) dx = +\infty$

• Do you mean such that $\displaystyle \lim _{n \to \infty} f_n(x)=0$ and $\displaystyle \lim_{n \to \infty} \int _0 ^ 1 f_n(x) dx = +\infty$? – Rory Daulton Jun 10 '15 at 12:47

$$f_n(x) = \begin{cases} 6n^3x(1-nx), & 0\le x\le \frac 1n \\ 0, & \frac 1n<x\le 1 \end{cases}$$

If you dislike definitions by cases, you could rewrite this as

$$f_n(x)=\max\left(6n^3x(1-nx),0\right)$$

Then for any given $x$, $f_n(x)=0$ if $n$ is large enough, but $\int_0^1 f_n(x)\, dx = n$ for all $n$.

Here are $y=f_1(x)$ through $y=f_{6}(x)$. (Ignore the values outside $x\in[0,1]$.) Note that the maximum value of $f_n(x)$ is $\frac 32n^2$ and occurs at $x=\frac 1{2n}$.

What about $f_n (x) = n^3 x \exp(-nx)$? $f_n(0) = 0 \to 0$ and for $x >0$ we have $\exp(-nx) \to 0$ much faster than $n^3 \to +\infty$, so $f_n(x) \to 0 \, \forall x$. But by substituting $y=nx$:

$$\int_0^1 n^3 x \exp(-nx)\, dx = n \int_0^n y \exp(-y) \, dy \to +\infty$$

You can take the piecewise linear function $f_n(x), n\ge 2$ such that $f_n(0)=0$ , $f_n(\frac{1}{n}) = n^2$ , $f_n\left(\frac{2}{n}\right)=0$ and $f_n(1)=0$.

Then you have $\forall x \in [0,1], \lim\limits_{n \to \infty} f_n(x)=0$ and $\lim\limits_{n \to \infty} \int\limits_{0}^1 f_n(x) dx = \lim\limits_{n \to \infty} n = +\infty$.

To see how this sequence of functions behaves, you can take a look at this animation : https://www.desmos.com/calculator/l0ap9b58qr .

this might be the right answer: $$f_n(x)=-1/(nx^3).$$

• $-\frac 1 n x^{-3}$ is not continuous on the domain $[0,1]$ – user228113 Jun 10 '15 at 13:26

$f_n(x)= n^3x^n(1-x)$ will do the job.