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Let us have a sequence of functions $\{f^K\}_{K \in \mathbb{N}} \in C([0,1],\mathbb{R})$ which is uniformly bounded in $L^2((0,1))$. We observe a sequence of Riemann sums $$R^K=\frac{1}{K} \sum_{k=0}^{K-1} |f^K((k+\frac12)/K)|^2.$$ Let $R^K$ be uniformly bounded. Can we claim $$\liminf_{K \to \infty} R^K \geq \int_0^1 |f(t)|^2~ dt,$$ where $f \in L^2((0,1))$ is the weak limit of (a subsequence of) $f^K$?

This seems as a wild guess without additional assumptions ("equiconvergence" of Riemann sums to the corresponding integrals), but can't find a counterexample. Any ideas or references?

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No. E.g., for natural $K\ge2$, let $$f^K:=1-g^K,$$ where, for $t\in[0,1]$, $$g^K(t):=\sum_{j=0}^{K-1}\Big(1-K^2\Big|t-\frac{j+1/2}K\Big|\Big)_+$$ and $u_+:=\max(0,u)$. That is, for each $t\in[0,1]$, we have $g^K(t)=1-K^2\big|t-\frac{j+1/2}K\big|$ if there is some $j\in\{0,\dots,K-1\}$ such that $\big|t-\frac{j+1/2}K\big|\le\frac1{K^2}$, and $g^K(t)=0$ otherwise. Here is the graph of $f^{10}\,$:

enter image description here

Then $f^K(\frac{k+1/2}K)=0$ for all $k\in\{0,\dots,K-1\}$ and hence $R^K=0$. On the other hand, $0\le g^K\le1$ and hence $$\int_0^1|g^K(t)|^2\,dt\le\int_0^1 g^K(t) \,dt=\frac1K\to0$$ (as $K\to\infty$), so that $g^K\to0$ and $f^K\to1$ strongly and hence weakly. Thus, $$\liminf_{K \to \infty} R^K=0 \not\ge1= \int_0^1 |f(t)|^2\, dt.$$

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