Find examples of sequences of functions in the space $L^\infty\left([0,1],\mathcal B\left([0,1]\right),\lambda\right)$ s. t. $\lim\limits_{n\to\infty}\|f_n-f\|_2=0$ and $\lim\limits_{n\to\infty}\|g_n-g\|_2=0,$ but $$\|f_ng_n-fg\|_2\not\to 0.$$ Is it possible to find such examples in the measure space $(\Bbb N,\mathcal P(\Bbb N),\mu),$ where $\mu$ is the counting measure?
My attempt:
Consider the sequence of functions $(f_n)_{n\in\Bbb N},f_n:[0,1]\to\Bbb R,$ defined by $$f_n(x):=x^p1_{\left[\frac1n,1\right]}(x),$$ where $p\in\left(-\frac12,-\frac14\right]$ and $f:[0,1]\to\Bbb R,$ $$f(x)=0\cdot 1_{\{0\}}+x^p1_{(0,1]}.$$ Then $\|f_n\|=\frac1{n^p}<+\infty,n\in\mathbb N$ so $f_n\in L^\infty\left([0,1],\mathcal B\left([0,1]\right),\lambda\right)$ and $$\begin{aligned}\lim_{n\to\infty}\|f_n-f\|_2&=\lim_{n\to\infty}\left(\int_{[0,1]}|f_n-f|^2d\lambda\right)^{\frac12}\\&=\lim_{n\to\infty}\left(\int_{[0,1]}\left|x^p1_{\left[\frac1n,1\right]}(x)-x^p1_{(0,1]}(x)\right|^2d\lambda(x)\right)^{\frac12}\\&=\lim_{n\to\infty}\left(\int_{[0,1]}\left|x^p1_{\left(0,\frac1n\right]}(x)\right|^2d\lambda(x)\right)^{\frac12}\\&=\lim_{n\to\infty}\left(\int_{\left(0,\frac1n\right)}x^{2p}d\lambda(x)\right)^{\frac12}\\&=\left(\lim_{n\to\infty}\int_0^{\frac1n}\frac{x^{2p+1}}{2p+1}\right)^{\frac12}=0.\end{aligned}$$ With the same computations I got $$\begin{aligned}\lim_{n\to\infty}\|f_n^2-f^2\|_2&=\lim_{n\to\infty}\lim_{\varepsilon\to 0^+}\left(\ln x\bigg|_\varepsilon^{\frac1n}\right)^{\frac12},\end{aligned}$$ so, I think, the limit does not exist.
I'm not quite sure this is right, but in case it is, is it possible to find sequences of essentially bounded functions $(f_n)_{n\in\Bbb N},(g_n)_{n\in\Bbb N}$ satisfying the conditions of the task with their respective limits being essentially bounded?
Regarding the measure space $(\Bbb N,\mathcal P(\Bbb N),\mu),$ I believe there are no such examples, but I can't prove my claim. I know that empty set is the only set in whole $\mathcal P(\Bbb N)$ which has a zero $\mu$ measure and that, therefore, all essentially bounded functions on that space are necessarily bounded, but I couldn't find a way to incorporate that fact into my attempt.
Let $H_n$ be the $n$-th harmonic number, that is $H_n = \sum_{k=1}^{n} \frac{1}{k}$. Define the function $f_n$ on $[0,1]$ as the characteristic function of the set $[H_n,H_{n+1}] \bmod 1$ multiplied by $\sqrt[4]{n+1}$. Then $\|f_n\|_2 = 1/\sqrt{n+1}$, and $(f_n)$ converges to $0$ in $L^2$. However, for $g_n = f_n$ we have $\|f_n g_n\|_2=\|f_n^2\|_2 = 1$, hence $(f_ng_n)$ does not converge to $0$.