Find a sequence of continuous functions $f_n:[0,1] \rightarrow \mathbb{R}$ with the properties:

1) $\limsup f_n(x)=1\;$ and $\;\liminf f_n(x)=0\;\;\forall x \in [0,1]$.

2) $ \lim_{n \rightarrow \infty}\int_{[0,1]}f_n=\infty$.

In this question I don't want to find such a function in a closed form. Instead I want to construct it an describe it geometrically.

Can someone help me with a hint?

Thank you in advance!

  • $\begingroup$ For part 2 do you mean that $\lim_{n\to \infty} \int_{[0,1]} f_n =\infty$? $\endgroup$
    – Matt
    Nov 30, 2016 at 19:04
  • $\begingroup$ yes indeed! i wil edit it better $\endgroup$ Nov 30, 2016 at 19:07
  • $\begingroup$ This sounds like a great exam problem! Can you construct a single sequence of numbers $x_i$ with $\limsup x_n = 1$ and $\liminf x_n = 0$ as a way to get started? $\endgroup$ Nov 30, 2016 at 20:00
  • $\begingroup$ yes i already have constructed a sequence of functions with limsup 1 and liminf 0 and the limit of integrals is zero $\endgroup$ Nov 30, 2016 at 20:05
  • $\begingroup$ i find a little difficulty in the case whre the integral diverges to infinity $\endgroup$ Nov 30, 2016 at 20:06

1 Answer 1


Here's a "geometric" idea: Choose a strictly increasing sequence $a_n$ in $(0,1)$ with $a_n\to 1.$ For each $n,$ put a triangle of height $n/(a_{n+1}- a_n)$ over the base $[a_n,a_{n+1}].$ That gives you $f_n$ on $[a_n,a_{n+1}];$ define $f_n = 0$ everywhere else. Then $f_n$ is continuous and $f_n\to $ pointwise on $[0,1],$ while $\int_0^1f_n = n/2 \to \infty.$ The sequence $f_1, f_2 + 1, f_3, f_4 + 1, \dots$ will then provide a solution.

I know you said you didn't want a closed form solution, but I can't help myself since it's pretty simple: Define $f_n(x) = n^3x^n(1-x).$ Then $f_n\to 0$ pointwise on $[0,1],$ and $\int_0^1 f_n \sim n \to \infty.$ As above, $f_1, f_2 + 1, f_3, f_4 + 1, \dots$ then gives a solution.

  • $\begingroup$ thats a nice answer but i have to construct a gewmetric example $\endgroup$ Nov 30, 2016 at 20:19
  • $\begingroup$ See my comment above to reduce your problem to a simpler one. $\endgroup$
    – zhw.
    Nov 30, 2016 at 20:20
  • $\begingroup$ @capo I added a "geometric" solution to my answer. $\endgroup$
    – zhw.
    Nov 30, 2016 at 20:43

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