Does convergence in $L^p$ imply convergence in distribution? 
$f_1,f_2,\dots ,$ and $f$ are in $L_{loc}^1(U)$.
Prove that if
$f_n\in L^p(U)(1\leq p\leq \infty)$ and $f_n\to f$ in the $L^p$ norm or weakly in $L^p$,
then $f_n\to f$ in $\mathcal{D}^\prime (U)$,
where $\mathcal{D}^\prime (U)$ denotes the space of all distributions on $U$.

I'm really stuck and I would appreciate if you indicate what aspect of $L^p$ convergence I should look at to tackle this problem. Any hints or suggestions would help.
Thank you.
 A: $f_n \to f$ in $D'$ if
$$\forall \phi \in D, \ \lim_n \int_U (f-f_n)\phi = 0$$
If the convergence is strong in $L^p$, Hölder inequality gives us the result :
$$\left|\int_U (f-f_n)\phi \right| \leq \int_U \left|(f-f_n)\phi \right| \leq \|f_n-f\|_p\|\phi\|_q \to 0$$
If the convergence is weak, as $D\subset L_q$, the answer is immediate
A: Take a test function $\phi\in D(U)$ and consider the $f_n$ as distributions:
$$\begin{align}|\langle f_n,\phi\rangle-\langle f ,\phi\rangle|&=\left|\int_U\left(f_n(x)-f(x)\right)\phi(x)dx\right|\\
&\le   \int_U |f_n(x)-f(x)|\cdot|\phi(x)|dx\\
&\le \|f_n-f\|_{L^p(U)}\cdot\|\phi\|_{L^q(U)}\end{align}$$
with $\frac 1p+\frac 1q=1$ for Hölder's inequality. Therefore, if $f_n\to f$ in $L^p$, then the above difference converges to zero for any test function and hence we have the convergence in $D'(U)$.
Weak convergence is treated similarly: as $\phi\in L^p$, the integral $$\int_U\left(f_n(x)-f(x)\right)\phi(x)dx$$ converges to zero and hence we also have the convergence in $D'(U)$.
