Let $f_n$ be a sequence of continuous functions on $[0,1]$, such that $0 \le f_n \le 1$ and $$\lim_{n\to\infty}\int_{0}^{1} f_n(x)\,\mathrm dx=0.$$

a) The sequence ${f_n (x)}$ converges uniformly.

b) The sequence ${f_n (x)}$ converges for some $x$.

c) The sequence ${f_n (x)}$ might not converge for any $x$.

d) The sequence ${f_n (x)}$ converges for every $x$ but not necessarily uniformly.

The answer is c? I have no counter example in my hand.

  • $\begingroup$ For a), take $f_n(x)=x^n$ $\endgroup$ – Jean-Claude Arbaut Jun 21 '13 at 17:28
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    $\begingroup$ for c) $\mathbb{1}_{[k/n,(k+1)/n]}, 0\leq k<n, n=1,2,3,...$ (edit appropriately for continuity) $\endgroup$ – yoyo Jun 21 '13 at 17:28
  • $\begingroup$ @yoyo You will have to make that a triangle to satisfy the continuity restriction. $\endgroup$ – Lord_Farin Jun 21 '13 at 17:29
  • $\begingroup$ @yoyo Is that continuous? $\endgroup$ – Pedro Tamaroff Jun 21 '13 at 17:30
  • $\begingroup$ @Lord_Farin, yeah, and to take $k$ the right way, but it does work. Hence b) and d) are wrong, and we're done. $\endgroup$ – Jean-Claude Arbaut Jun 21 '13 at 17:30

Consider the following sequence of sets, constructed as follows. Let $I_1=[0,1]$. Split $[0,1]$ in half, and let $I_2$ and $I_3$ be the first and second halves. Split $[0,1]$ in fourths, and let $I_4,I_5,I_6,I_7$ be the fourths in order. Continue this way, split $[0,1]$ in $2^n$ths and let $I_{2^n},\cdots, I_{2^{n+1}-1}$ be the $2^n$ pieces in order.

Now let $$f_n(x)=\text{ an isosceles triangle over } I_n \text{ of height } 1$$

Then $$\int_0^1 f_n\to 0$$ but if $x\neq m2^{-k}$ then $$\tag 1 \limsup_{n\to\infty} f_n(x)=1 \; ; \;\liminf_{n\to\infty} f_n(x)=0$$

ADD To fix the above, we can do the following: let $\langle g_n\rangle $ be the same as $\langle f_n\rangle $, but with the triangles shifted so that the peaks of the triangles fall on the binary rational points. Thus, $g_1$ will look like (the right) half an isosceles triangle on $[0,1/2]$ plus (the left) half of an isosceles triangle on $[3/4,1]$, $g_2$ will be an isosceles triangle over $[1/2,3/4]$, $g_3$ will be the two halves at $[0,1/4]$,$[7/8,1]$, and so on. Then $\langle g_1,f_1,g_2,f_2,g_3,f_3,\ldots\rangle$ should do: now the rational binary points oscillate between $0$ and $1$.

  • $\begingroup$ @arbautjc I like to imagine the sequence as a series of lights that light up from left to right in a nice way. Seems more entertaining. $\endgroup$ – Pedro Tamaroff Jun 21 '13 at 17:36
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    $\begingroup$ Yes, but I was awfully wrong: it does not prove that the function may not converge for any value. Silly me :-) $\endgroup$ – Jean-Claude Arbaut Jun 21 '13 at 17:38
  • $\begingroup$ @arbautjc True. $\endgroup$ – Pedro Tamaroff Jun 21 '13 at 17:38
  • $\begingroup$ @Peter: I would add additional triangles in the middle, for example between the triangle $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ insert $[\frac{1}{4},\frac{3}{4}]$. (Perhaps there is a better way to do it, I don't know.) Probably with judicious use of triangles you can assure the function may not converge anywhere. $\endgroup$ – sdcvvc Jun 21 '13 at 18:57
  • $\begingroup$ @sdcvvc Oh, I see. Let me try. $\endgroup$ – Pedro Tamaroff Jun 21 '13 at 19:03

How about this:

$f_1=1$ everywhere;

$f_2=1$ on the left half of $[0,1]$ and $0$ on the right half; $f_3=1$ on the right half and $0$ on the left;

$f_4$, $f_5$, $f_6$ are respectively equal to $1$ on the left third, the middle third, and the right third and $0$ elsewhere;

$f_7$, $f_8$, $f_9$, $f_{10}$ are equal to $1$ respectively on the first, second, third, and fourth quarters, and $0$ elsewhere;

and so on.

Then $f_n(x)$ diverges for every $x$ as $n\to\infty$, but $\displaystyle\int_0^1 f_n\to0$.

  • $\begingroup$ But that is not continuous. That is the example I had in mind at first. When making it continuous with triangles instead of indicator functions, we loose the property that the sequence is nowhere convergent. $\endgroup$ – Pedro Tamaroff Jun 21 '13 at 21:29
  • $\begingroup$ @PeterTamaroff that's easy to fix: Just take $f_2$ to be $0$ on $[0,\tfrac12-\epsilon]$ and $1$ on $[\tfrac12,1]$ and join the graph up so it's piecewise linear. You'll still have that $f_n(x) = 1$ frequently for every $x \in [0,1]$ and that $\int_0^1 f_n(x) \,dx \to 0$. $\endgroup$ – kahen Jun 21 '13 at 21:57
  • $\begingroup$ @kahen Awesome. I tried something like that. I ended up with a beautiful mess. $\endgroup$ – Pedro Tamaroff Jun 21 '13 at 23:33

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