# Does $L^2$ convergence imply that the pointwise limit is the most frequent accumulation point?

The typical example for a function that converges in $$L^2([0,1])$$ but not pointwise is the indicator function with width $$1/n$$ that wanders accross, the interval, i.e.

$$h_1 = 1_{[0, \frac{1}{2}]}, h_2 = 1_{[\frac{1}{2}, 1]}, h_3 = 1_{[0, \frac{1}{3}]}, 1_{[\frac{1}{3}, \frac{2}{3}]}, \ldots.$$

Then $$h_n \to 0$$ in $$L^2$$ but not pointwise.

And if I understand right, this is also the only way in which pointwise convergence can fail for an $$L^2$$ converging function: The $$h_n$$ do not converge to $$0$$ in the supremum norm and the place at which the supremum is attained happens wanders around, whereas the area in which the supremum is attained grows smaller.

But if we where to look at any single point evaluation $$h_n(x)$$ we could easily "read off" the right pointwise limit. The sequence would have multiple accumulation points (0 and 1) and the most common one (in this case 0) would be the limit.

My question is: Is this the case in general? If we have an $$L^2$$ converging function, we can just check the point evaluations and we would be able to easily "read off" what the pointwise limit should be by taking the dominant accumulation point?

I am interested in this because of stochastic analysis. Here you can define the stochastic integral only as an $$L^2$$ limit, not a pointwise limit.

• Point evaluation of $L^{2}$ functions does not make any sense because elements of $L^{2}$ are not ordinary functions. Mar 6 at 23:49