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While working through problems, I came upon one that I couldn't figure out and was wondering how to do. Here's the problem:

Find a sequence $\{f_n\}^\infty_{n=1}$ of nonnegative measurable functions such that $\displaystyle\lim_{n\to\infty} \int f_n = 0,$ but $\displaystyle\limsup_{n\to\infty} f_n(x) = + \infty$ for all $x.$

Is this a classical example? I've never seen it before if so...

Edit: the domain of integration is $\mathbb{R}$

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  • $\begingroup$ for all $x\in \mathbb{R}.$ I think the idea is to take something like the sequence $f_0, f_1, \ldots$, where $f_{n(n-1)/2 + j}$ is the characteristic function on $[j/n, (j+1)/n].$ Obviously some modification must be made for this to work though... and I'm not seeing how to extend it to all of the reals $\endgroup$ – cats Mar 1 '13 at 10:12
  • $\begingroup$ edited my answer. Hope the new one helps $\endgroup$ – Quickbeam2k1 Mar 1 '13 at 16:39
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So I found a different simpler idea. The easiest way is to use two indices. Assume first $x\geq 0$.

Basically, the idea is to split the interval $[0,2^k]$ into $2^{4k}$ pieces of height $2^k$ and width $2^{-2k}$.

Define the function $$f_{k,l}(x):=2^k\chi_{[l2^{-2k},(l+1)2^{-2k}]}(x)$$ and think of $0\leq l \leq 2^{4k}-1$. Obviously for $k,l$ fixed the Integral is $2^k 2^{-2k}=2^{-k}$ hence $\int f_{k,l}\to 0$ for $k\to\infty$.

For fixed $k$ we see that $$[0,2^k]=\bigcup_{l=1}^{2^{4k}-1} [l2^{-2k},(l+1)2^{-2k}].$$ Thus for every point in $x\in[0,2^k]$ there exist an $l\in (0,2^{4k}-1)$ and hence there is an $l$ such that $f_{k,l}(x)=2^k$.

Now all we have to do is define $g_n$ appropriately in terms of $k,l$. For example. After increasing $k$ by one we have to go through all the $l$ values from $0$ to $2^{4k}-1$. Then increase $k$ again and so forth.

Finally you can choose the sequence $h_n(x)=g_n(x)+g_n(-x)$ to cover all points of $\mathbb{R}$

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  • $\begingroup$ I want $\limsup f_n(x) = +\infty$ everywhere though. $\endgroup$ – cats Mar 1 '13 at 9:48
  • $\begingroup$ To me it seems that $\limsup_{n\to\infty} f_{n}(x)=0$ for all $x\neq 0$ in this example. $\endgroup$ – T. Eskin Mar 1 '13 at 9:49
  • $\begingroup$ thanks, updated my idea :) $\endgroup$ – Quickbeam2k1 Mar 1 '13 at 9:51
  • $\begingroup$ Why does this $g_{n}$ do the job? You only get $+\infty$ with $f_{n}$ in the origin, and everywhere else $0$. How do you enumerate $\mathbb{Q}$ so that for any $x\in\mathbb{R}$ and $\varepsilon>0$ you find $k\in\mathbb{N}$ with $|q_{n}+x|<\varepsilon$ for all $n\geq k$. You can't have $0$ appearing infinitely often in the sequence $(q_{n}+x)_{n=1}^{\infty}$ for every $x\in\mathbb{R}$, so I don't really see why this example works. Could you elaborate more on it? $\endgroup$ – T. Eskin Mar 1 '13 at 9:56
  • $\begingroup$ i was actually just about to say that; you don't know that every real appears infinitely many times. $\endgroup$ – cats Mar 1 '13 at 9:57

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