A basic question on the space of square integrable functions

Question

I have seen in a book the following cliam:

Let $f_m,f \in L^2[0,N]$

and $\frac{1}{m}\sum_{k=1}^{m}f_{n(k)} \to f$ in $L^2[0,N]$ for a subsequence $n(k)$

Then for any $g \in L^2[0,N] s.t. \|g\|=1$ $$\int_{0}^{N}g(x)f_m(x) \to \int_{0}^{N}g(x)f(x)$$.

Answer 1

This is not correct. Let $f_m = 1$ if $m$ is even and $f_m = 0$ if $m$ is odd. Then

$$\frac{1}{m} \sum_{k=1}^m f_m \to \frac 12$$

in $L^2[0, N]$. But for $g(x) = 1$,

$$\int_0^N g(x) f_m(x) dx = \int_0^N f_m(x)dx = N \text{ or }0$$

does not converge.

Answer 2

What is correct is the following:

If The sequence $(f_m)_m$ is bounded and if furthermore $\frac{1}{m} \sum_{k=1}^m f_{n_k} \to f$ in $L^2$ holds for every subsequence $n(k)$ (or even if every subsequence $n_k$ has a further subsequence $n_{k_l}$ such that $\frac{1}{m} \sum_{l=1}^m f_{n_{k_l}} \to f$ holds), then your claim is true.

To see this, assume that the claim is false. Then there is some $g\in L^2$ such that $\int g f_m \to \int g f$ does not hold. This yields some $\epsilon >0$ as well as a subsequence $(f_{n(k)})_k$ such that

$$ \bigg|\int g [f_{n(k)} - f] \bigg| \geq \epsilon \qquad (\dagger) $$

for all $k$.

But as a bounded sequence in the Hilbert space $L^2$, the sequence $(f_{n_k})_k$ has a weakly convergent subsequence $(f_{n(k(l))})_l$. Let the weak limit of this sequence be $h$. By assumption, we have $\frac{1}{m} \sum_{l=1}^m f_{n(k(l))} \to f$ with convergence in $L^2$. But the same sequence also converges weakly to $h$ (why exactly?), so that $h=f$.

But this implies

$$ \int g f_{n(k(l))} \to \int g h =\int g f, $$ in contradiction to $(\dagger)$ above.