weak convergence in $L^2$ / $C$ ==> pointwise convergence I have a sequence of function $B_n \in C([0,1],R)$ and a $B \in C([0,1],R)$, such that:
$\int_{[0,1]} B_n(x) f(x) dx \rightarrow \int_{[0,1]} B (x) f(x) dx$ for all $f \in C([0,1],R)$ which are bounded.
Does this imply point wise convergence of $B_n(x) \rightarrow B(x)$ for all $x$?
My starting idea:
This convergence implies weak convergence in $L^2([0,1])$, right? 
Do I have here a result for point wise convergence? 
 A: The standard counterexample is the trigonometric system. Let $B_n(x)=\sin(n\,\pi\,x)$. By the Riemann-Lebesgue lemma $B_n$ converges weakly to $0$ in $L^2$, but $B_n$ does not converge pointwise.
A: Your condition doesn't imply weak convergence in $L^2$ either.  Let $B_n$ have a "spike" of area 1 in the interval $(2^{-(2n+2)},2^{-(2n+1)})$, a negative spike of area 1 in $(2^{-(2n+1)}, 2^{-2n})$, and vanish elsewhere.   Then for continuous $f$, $\int B_n f$ looks approximately like $f(2^{-(2n+2)})-f(2^{-(2n+1)})$ which goes to 0 as $n \to \infty$.  But we could pick $f \in L^2$ which oscillates near 0 in such a way that $\int B_n f$ does not converge; something like $\sin(2^{-1/x})$.
The moral is that it isn't sufficient to check weak convergence on a dense set.
A: Unless I am misunderstanding the usage of "R," the answer is no. Take a $B_n$ that converges to 0 in measure. The typical example of this is a continuous function  approximating a step function of width $1/n$ and height 1, which slides back and forth on $[0,1]$. This converges to 0 in measure, and 0 is definitely continuous. It does not converge pointwise, since the sliding peak. Moreover, for each continuous function, the integrals converge since continuous functions are bounded on $[0,1]$ (use say, dominated convergence theorem). 
