Consider $\Omega=[0,1]$ equipped with Borel $\sigma$-algebra and Lebesgue measure. Consider the sequence $f_n = \sum_{i=1}^{2^n} (-1)^{i-1} 1_{\big[ \frac{i-1}{2^n}, \frac{i}{2^n} \big) }$ in $L^p$ where $1<p<+\infty.$ I believe that $f_n \rightarrow 0$ weakly in $L^p$ yet I am not able to show it.
We know by Banach-Alaoglu theorem that the closed unit ball in $L^p$ is weakly compact. In other words every bounded sequence has a weakly convergent sequence and the sequence $(f_n)_n$ is clearly bounded. At least we know that $$ f_{n_k} \rightarrow g \text{ (weakly)}$$ for some $g \in L^p.$ Yet I could not continue further. Are there more practical ways of varifying weak convergence rather than using the definition?
Any help is appreciated, thanks in advance!
Indeed, it is usually not an easy task to show that $$\tag{*}\int f_n h\to \int fh$$ for all $h\in L^{p'}$ in general, especially when we don't know a priori the limiting function. However, using the fact that $(f_n)_{n\geqslant 1}$ is bounded in $L^p$ and Hölder's inequality, it suffices to show, by approximation by step functions, that $(*)$ holds when $h$ is the indicator function of a Borel set. Using regularity of the Lebesgue measure, we can see that $f_n\to f$ weakly in $L^p$ if and only if (*) holds for $h=\mathbf 1_{O}$ where $O$ is an open subset of $[0,1]$. Since such sets can be writing as a countable disjoint union of open intervals, we derive that $f_n\to f$ weakly in $L^p$ if and only if $\sup_{n\geqslant 1}\lVert f_n\rVert_p$ is finite and $\int_I f_n \to \int_I f$ for all interval $I$.
Now, using density of $\{k2^{-N},N\geqslant 1, 0\leqslant k\leqslant 2^N\}$, it suffices to consider intervals $I$ whose end points are $k2^{-N}$ and $k'2^{-N}$ respectively. In this case, one can compute explicitely $\int_{(k2^{-N},k'2^{-N})}f$.