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.

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, $$\liminf_{K \to \infty} R^K=0 \not\ge1= \int_0^1 |f(t)|^2\, dt.$$

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 $f^{10}\,$:

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, $$\liminf_{K \to \infty} R^K=0 \not\ge1= \int_0^1 |f(t)|^2\, dt.$$

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.

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, $$\liminf_{K \to \infty} R^K=0 \not\ge1= \int_0^1 |f(t)|^2\, dt.$$

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.

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, $$\liminf_{K \to \infty} R^K=0 \not\ge1= \int_0^1 |f(t)|^2\, dt.$$