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I'm looking for a sequence of functions that is continuous and absolutely integrable, but pointwise divergent for every $z$ $\in [0,1]$.

In other words,

$ \int_0^1 |f_n(z)| dz \rightarrow 0$ as $n \rightarrow \infty$, but $(f_n(z))_n$ pointwise diverges for every $z$ $\in [0,1]$.

I'm having no luck coming up with an example, and would appreciate any sort of insight on how to go about constructing such a sequence.

Eddie
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    What do you mean by "diverges"? Do you just mean that $\lim_{n\to\infty} f_n(z)$ doesn't exist, or that $\lim_{n\to\infty} f_n(z)=+\infty$? – Math1000 May 11 '15 at 02:25
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    I guess diverges would mean "not converge" to a finite number . @Math1000 –  May 11 '15 at 02:28
  • @Math1000 John is right, that's what I mean. – Eddie May 11 '15 at 02:31

1 Answers1

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For integers $n > 0$ and $0 \leq m < n$, Define $f_n^m(z) = 1$ if $z \in [m/n,(m+1)/n]$ (call this interval $I_n^m$), and $0$ otherwise. Then "smooth the edges" (to make them continuous) of these functions so that they're not quite step functions, and arrange them in the sequence $$f_1^0,f_2^0,f_2^1,f_3^0,f_3^1,f_3^2,f_4^0,f_4^1,f_4^2,f_4^3,\dots$$

Each of these diverges pointwise, because for any $x \in [0,1]$, you should be able to find two subsequences of ${f_n^m(x)}$, one that converges to $0$ and another that converges to $1$. But the integrals get smaller and smaller since $$\lim_{n\to\infty}\mu(I_n^m) = \lim_{n\to\infty} \frac 1 n = 0,$$ where $\mu([a,b]) = b-a$. Not quite a precise answer, but if you work out the details it should work.

Edit: If $J_n^m = \text{Interior}(I_n^m)$, then instead of taking step functions and smoothing them, you can just choose continuous $f_n^m$ such that $f_n^m = 0$ outside of $J_n^m$, and $f_n^m \equiv 1$ on $I_n^m \setminus (\text{tiny boundary interval})$ and the result should hold. Examples include triangles, Gaussians, trapezoids, etc.

Ashwin Iyengar
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