Recently in my class we discussed how for $\ 1<p<+\infty$ if a sequence $\{f_n\} \subset L^p$ converges in distribution to $f$ and is bounded, i.e. $||f_n||_p \leq C$ $\forall n$, then it also converges weakly in $L^p$. I've been trying to find counterexapmlpes to this fact, both when the bound condition is satisfied for a general $p > 1$, and for $p = 1$ with a bounded sequence, though I always run into this kind of trouble:

take for example the sequence $f_n(x) = n(\chi_{(0,\frac{1}{n})} - \chi_{(-\frac{1}{n},0)})$. It is clearly bounded in $L^1(\Bbb R)$, since $||f_n||_1 = 2$ for all $n$, though for a test function $\phi$ you get $$\int_\Bbb R f_n \phi dx = n\int_0^\frac1n\phi(x) dx - n\int_{-\frac1n}^0 \phi(x) dx = n\int_0^\frac1n (\phi(x) - \phi(-x) )dx$$ having changed $x \mapsto -x$ in the second integral. Now by Lagrange's theorem there is an $x_n \rightarrow 0$ such that $2x\phi'(x_n) = \phi(x) - \phi(-x)$, so the last espression equals

$$2n\phi'(x_n)\int_0^\frac1n xdx = \frac{\phi'(x_n)}n \rightarrow 0$$ Clearly if my calculations are right we are missing an extra $n$ in front of the integral, so that we can get the desired result of $\phi'(0)$. Doing this though we lose the boundedness condition since $||nf_n||_1 = 2n \rightarrow \infty$. I've expreimented with many similar sequences, and they all seem to have the same problem. So I ask you:

1)Am I doing things right here?

2)Can you give me some exalmpes that actually work of sequences converging in distribution but not weakly in $L^p$ (especially for $p=1$)?

I'm also interested in not bounded ones though I think that $nf_n$ is already a good example since tested with $g(x) = \chi_{(-1,1)}(x) \in L^q(\Bbb R)$ I get $\int nf_n(x)g(x)dx = n \rightarrow \infty$.


First, we have to show that $f$ belongs to $\mathbb L^p$. To this aim, extract a weakly convergent subsequence $\left(f_{n_j}\right)_{j\geqslant 1}$ of $\left(f_{n}\right)_{n\geqslant 1}$, say to $g$. Using the convergence in distribution and weakly of the subsequence, we get that for all $\phi\in\mathcal D\left(\mathbb R\right)$, $$\int_{\mathbb R}\left(f(x)-g(x)\right)\phi(x)\mathrm dx=0$$ hence $f=g$. By definition of weak convergence, the function $g$ belongs to $\mathbb L^p$, hence so does $f$.

Now, the weak convergence of $\left(f_{n}\right)_{n\geqslant 1}$ to $f$ follows from its boundedness in $\mathbb L^p$ and density of the set of test functions in $\mathbb L^p$.

  • 1
    $\begingroup$ Sorry but his doesn't answer my question. I asked for a bounded sequence that does NOT converge weakly in $L^1$, while converging in distribution. $\endgroup$ – barmanthewise Jan 18 '18 at 20:14
  • $\begingroup$ You are right. I will try to think about it. $\endgroup$ – Davide Giraudo Mar 13 '18 at 12:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.