# Convergence in $L_1$ counterexample

I am looking for an example of a sequence of r.v. $$X_n$$ that converges to $$X$$ in $$L_1$$, but such that $$X_n^2$$ does not converge to $$X^2$$ in $$L_1$$.

Anyone has something in mind?

• I’m not sure why this has to be about random variables; it sounds like a real analysis question. There’s an answer on our sister site: math.stackexchange.com/questions/811765/… – Arya McCarthy May 31 at 14:54
• @AryaMcCarthy it's not evident for most the relation between real analysis and probability. – Manuel May 31 at 15:22

Let $$X_n \sim Be(n^{-1})$$ that is a Bernoulli random variable. Now consider $$Y_n = \sqrt{n}X_n$$.

It is straight forward that $$E(|Y_n-0|)= n^{-1/2}$$. Hence $$Y_n \overset{L_1}{\to} 0$$.

Since $$E(|Y_n^2 - 0^2|) = 1$$ you get that $$Y_n^2 \not \overset{L_1}{\to} 0^2$$

As a side note, this examples shows that $$L_1$$ convergence is not preserved by contiuous transformations i.e if $$g : \mathbb R \to \mathbb R$$ is a continuous function then $$X_n \overset{L_1}{\to} X \quad \not \Rightarrow \quad g(X_n) \overset{L_1}{\to} g(X)$$

Convergence in distribution, probability and and almost everywhere are all proserved by continuous transformations.

• Could you also give $f_n = n^{-1}1_{[0,n]}$? That way it converges to $||f_n||_{1}=1$ but when you take the square it does not converge at all. – Ariel May 31 at 15:32
• @Ariel it is not clear to me how is $f_n$ defined as a random variable. – Manuel May 31 at 15:59
• Since we just need it to be a $\mathcal{F}$-measurable function with some work I think we could think of it as r.v.s for given $n$s. Mostly though I just wanted to check my intuition. I think your earlier comment is correct though and $f_n$ does not work because it does not converge in L1! (I should have drawn a picture :) ) – Ariel May 31 at 16:31
• Thank you! this is what I was looking for – Rayan Mezher Jun 1 at 7:50
• @RayanMezher you can accept the answer if you consider so. – Manuel Jun 1 at 18:23

In the answer of the linked question over on Math.SE and a comment of this page, it is suggested to take $$f_n = n^{-1} \mathbf 1_{[0,n]}$$ actually this does not work, and this is because this example solves the converse problem ($$L^2$$ but not $$L^1$$), and further on a space that is not a probability space ($$\mathbb R$$ with uniform measure). Note that $$\|f_n\|_{L^1} \equiv 1$$, which shows that the $$L^1$$ norm converges, but this does not give you convergence in $$L^1$$ norm, as the almost everywhere limit is $$f=0$$, and if $$f_n$$ did converge in $$L^1$$ then it would have to converge to the a.e. limit i.e. $$\|f_n - 0 \|_{L^1} \to 0$$. The $$L^2$$ norm however, converges to $$0$$, proving convergence in $$L^2$$ norm to zero.

(In fact there is even a slight discrepancy in the question, as the difference in $$L^2$$ can be written $$\|f_n - f\|^2_{L^2} = \|(f_n-f)^2\|_{L^1}$$ but convergence of the square in $$L^1$$ is $$\|f_n^2 - f^2\|_{L^1}$$. Thankfully this nonlinearity issue disappears when the $$f$$ is zero.)

A correct example was given in the math.SE question's body: $$\sqrt n\mathbf 1_{[0,1/n]}$$ (with uniform probability on $$[0,1]$$). Another perhaps more trivial (and perhaps not in the spirit of the question) example can be given via constant (in $$n$$) sequences, simply because there are functions in $$L^1$$ whose square are not in $$L^1$$, so the question of their convergence has no meaning. Any random variable with a mean and without variance will do; an example using the same uniform probabilty on $$[0,1]$$ is $$\frac1{\sqrt x}$$.

PS One can prove partial results. For example, if $$f,f_n$$ are almost surely bounded uniformly in $$n$$, $$\|f\|_{L^\infty} , \|f_n\|_{L^\infty} \le M$$, then observe $$\|f_n^2 - f^2\|_{L^1} = \|(f_n - f)(f_n+f)\|_{L^1}\le M\| f_n - f\|_{L^1} \to 0.$$ This is of course consistent with the example of Manuel as his $$Y_n$$ is not a.s. bounded in $$n$$.

• I see, thank you! – Rayan Mezher Jun 1 at 7:49
• @Rayan you are welcome; by the way, the example here that j quote from the math.se page is actually the same example of Manuel in different language – Calvin Khor Jun 2 at 1:37