# Find a sequence converging in distribution but not weakly

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 $\mathbb L^p$.
• 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. – barmanthewise Jan 18 '18 at 20:14