# Does pointwise convergence imply convergence in distribution? Counterexample?

$f_1,f_2,\dots ,$ and $f$ are in $L_{loc}^1(U)$.

I'm trying to give a counterexample where $f_n\to f$ pointwise,

but not $f_n\to f$ in $\mathcal{D}^\prime (U)$,

where $\mathcal{D}^\prime (U)$ denotes the space of all distributions on $U$.

I'm trying to come up with a sequence of functions which converges pointwise but does not converge almost everywhere or in $L^p$, and then conclude it does not converge in distribution, for convergence in distribution looks so hard to deal with. But I couldn't find anything yet. I'm not sure if this approach is reasonable.

I would appreciate any comment or hint. Thank you for reading.

This is similar to the fact that pointwise convergence does not imply $L^1$ convergence.

Construct a sequence $f_n$, each with integral $1$, that converges to $0$ pointwise.

• Thank you,@Dunham. You mean like, $f_n (x)=\frac{1}{n}x^{n-1}$ on $(0,1)$ so that the integral of each function is $1$ and it converges to $0$ pointwise? Commented Apr 7, 2015 at 16:59
• You want $f_n(x)=nx^{n-1}.$
– zhw.
Commented Apr 7, 2015 at 17:05
• Thank you, for the correction, @zhw. Commented Apr 7, 2015 at 17:09
• Actually that doesn't work, because on $U=(0,1)$ these $f_n$'s converge to $0$ pointwise and in $D'(U).$ Let's try $U=\mathbb{R},$ with $f_n(x)=nx^n\chi_{(0,1)}.$ This sequence converges pointwise to $0$ everywhere, but $f_n \to \delta_0$ in $D'(\mathbb {R}).$
– zhw.
Commented Apr 7, 2015 at 17:13
• It follows from: $\int_0^1nx^ng(x)\,dx \to g(1)$ for any $g\in C([0,1]).$ This is an exercise you need to do if you're studying distributions.
– zhw.
Commented Apr 7, 2015 at 18:22