Weak convergence implies $L^p$ convergence on a smooth bounded domain? Let $U\subset \mathbb{R}^n$ be a bounded domain with smooth boundary. Let $f_k\in L^p(U)$. Does weak convergence of $f_k$ to $f \in L^p$ implies $L^p$-convergence of $f_k$ to $f$? By weak convergence I mean that 
$$
\int_U f_kg \,dx \rightarrow \int_U fg\, dx
$$
for all $g\in L^q(U)$ where $\frac{1}{p}+\frac{1}{q}=1$. 
 A: This is perhaps a sledgehammer approach, but the answer is "no" for $p>1$.
The closed unit sphere of a reflexive space is weakly compact. In fact, it is sequentially weakly compact by the Eberlein–Šmulian theorem. On the other hand, the closed unit sphere of an infinite dimensional space is not norm compact. From these facts, it follows that in an infinite dimensional reflexive Banach space, there exists a weakly convergent sequence of norm one elements that is not norm convergent.
(I'm fairly certain the answer is "no" for $p=1$ as well.)
A: Another standard one-dimensional example is the sequence of Rademacher functions $r_n(x)=\operatorname{sign}\sin (2^n\cdot 2\pi x)$ on $[0, 1]$. The $L^p$  norm of $r_n$ is equal to $1$ for every $n$ and every $p\in [1,\infty]$. 
Yet, the sequence converges weakly to $0$ in every $L^p$ for $p\in [1,\infty)$. Since $r_n$ are uniformly bounded, it suffices to show that $\int_0^1 r_n(x)g(x)\,dx\to 0$ for every $g\in C[0,1]$. (This works even when $p=1$, because every element of $L^\infty$ can be approximated by continuous functions in $L^1$ norm.) 
Using the uniform continuity of $g$, we estimate the integrand in $$\int_0^1 r_n g=\sum_{k=1}^{2^n}\int_{2^{-n}(k-1)}^{2^{-n}(k-1/2)} (g(x)-g(x+2^{-n-1})) \,dx \tag1$$ by $\epsilon=\epsilon(n)\to 0$. 
The example generalizes to higher dimensions by letting $f_n(x_1,\dots,x_d)=r_n(x_1)\chi_Q$ where $Q=[0,1]^d$ (which we may assume to be contained in $U$). 
