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Let $(f_n)_{n\ge 1}$ be a sequence of function that converges to some $f\in L^2(\mathbb{R})$, i.e., $||f_n-f||_{L^2(\mathbb{R})}=0$.

Clearly the sequence of norms $(||f_n||_{L^2(\mathbb{R})})_{n\ge 1}$ is bounded. Moreover, it is well-known that $(f_n)$ has a convergent subsequence that converges almost surely.

I would like to prove/disprove the following statement: There exists $g\in L^2(\mathbb{R})$ such that $|f_n|\le g$ a.s. for all $n\ge1$.

Edit: As S.L. commented below, this statement is false if we endow $\mathbb{R}$ with the Lebesgue measure. I'm still wondering whether the above statement holds when we consider a probability measure on $\mathbb{R}$, for example, the Gaussian measure $\mu(A)=\int_A e^{-x^2/2}dx$.

StubbornAtom
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user107678
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  • This is a partial converse of the dominated convergence theorem and you can find many questions on this topic on this site. E.g., https://math.stackexchange.com/questions/2834822/inverse-of-the-dominated-convergence-theorem https://math.stackexchange.com/questions/462312/about-the-converse-of-the-lebesgue-dominated-convergence – whpowell96 May 23 '23 at 14:17

1 Answers1

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Consider the sequence $1_{[0,1/2]},1_{[1/2,1]},1_{[0,1/4]},1_{[1/4,2/4]},...,1_{[7/4,2]},1_{[0,1/8]},1_{[1/8,2/8]},...,1_{[23/8,3]},1_{[0,1/16]},...$

So the intervals have length $1/2^n$ and cycle through to $n$, then they halve, and start again from 0, except with half the length and cycling up to $n+1$.

Since the interval length converges to 0, this sequence converges to 0 in $L^2$, but any $g$ that bounds the sequence almost surely has to be greater than 1 on $[0,\infty)$ almost surely, so $g$ cannot be in $L^2$.

Edit: I've had a question asking for a counterexample in the case of probability measures, so here it is:

Consider the space $([0,1],\mathcal{B}([0,1]),\mu)$ where $\mu$ is the lebsegue measure, and the sequence of random variables:

$1_{[0,1/2]},1_{[1/2,1]},2\cdot1_{[0,1/8]},...,2\cdot1_{[7/8,1]},4\cdot1_{[0,1/32]},...$

Now the $L^2$ norms are $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{4}},...,\frac{1}{\sqrt{4}},\frac{1}{\sqrt{8}},...$ which still tend to 0, but this any function that bounds this sequence needs to be greater than $n$ almost everywhere for every $n$.

The reason this still works is because, imagining the indicator functions on a graph, we can grow the 'height' (the coefficients) of the functions while shrinking the 'width' (interval length) so that the $L^2$ norm converges to 0.

Try it yourself for $L^P$, you just need to change the scale factors a bit.

S.L.
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  • If we consider a probability measure on $\mathbb{R}$, for example the Gaussian measure $\mu(A)=\int_A e^{-x^2/2}dx$ then we still have $g\in L^2(\mathbb{R},\mu)$. Can you give a counterexample in the case of probability measures? – user107678 May 24 '23 at 01:31