$L^2([-1, 1])$, we have $\lim_{j \to \infty} f_j(x) = 1$ for a.e. $x \in [-1, 1]$?

Question

Let$$\{f_j : j \in \mathbb{N}\} \subset L^2([-1, 1])$$be such that$$f_j \ge 0,\,\text{ }\|f_j \|_{L^1([-1, 1])} = 2,\,\text{ }\left|\|f_j\|_{L^2([-1, 1])} - \sqrt{2}\right| \le 2^{-j}.$$How do I see that$$\lim_{j \to \infty} f_j(x) = 1$$for a.e. $x \in [-1, 1]$?

Answer 1

Notice that

$$ \int_{-1}^{1} |f_j - 1|^2 = \| f_j \|_2^2 - 2\|f_j\|_1 + 2 = \mathcal{O}(2^{-j}). $$

By the monotone convergence theorem, we have

$$ \int_{-1}^{1} \sum_{j=1}^{\infty} |f_j - 1|^2 = \sum_{j=1}^{\infty} \int_{-1}^{1} |f_j - 1|^2 < \infty. $$

Therefore $\sum_{j=1}^{\infty} |f_j - 1|^2$ is finite a.e. and hence $f_j \to 1$ a.e.

Answer 2

I proved your statemement up to a subsequence $f_{j_k}$. We have \begin{align} \int_{-1}^1 |f_j(x)-1|^2dx&=\int_{-1}^1 f_j(x)^2dx -\int_{-1}^1 2f_j(x)dx +\int_{-1}^1 1dx \\ &=\| f_j\|_{L^2}^2-2\| f_j\|_{L^1}+2=\| f_j\|_{L^2}^2-2 \end{align} Since $\| f_j\|_{L^2}$ converges to $\sqrt 2 $ : $$\int_{-1}^1 |f_j(x)-1|^2dx\longrightarrow 0.$$ Now we can say that there is a subsequence $f_{j_k}$ such that $$\lim_{k \to \infty}f_{j_k}(x)=1$$ for a.e $x$ in $[-1,1]$.