0
$\begingroup$

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 $\displaystyle \lim_{n \to \infty} \int _0 ^ 1 f_n(x) dx = +\infty$

$\endgroup$
  • 4
    $\begingroup$ Do you mean such that $\displaystyle \lim _{n \to \infty} f_n(x)=0$ and $\displaystyle \lim_{n \to \infty} \int _0 ^ 1 f_n(x) dx = +\infty$? $\endgroup$ – Rory Daulton Jun 10 '15 at 12:47
4
$\begingroup$

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

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 $x=\frac 1{2n}$.

enter image description here

$\endgroup$
1
$\begingroup$

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 $y=nx$:

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

$\endgroup$
1
$\begingroup$

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 .

$\endgroup$
0
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.