IMC 1995 · Day 2 · P12

It is clear that one can add some functions, say $\{g_m\}$, which satisfy the hypothesis of the problem and the closure of the finite linear combinations of $\{f_n\} \cup \{g_m\}$ is $L_2[0,1]$. Therefore without loss of generality we assume that $\{f_n\}$ generates $L_2[0,1]$.

Let us suppose that there is a subsequence $\{n_k\}$ and a function $f$ such that $$f_{n_k}(x) \xrightarrow[k \to \infty]{} f(x) \quad \text{for every } x \in [0,1].$$ Fix $m \in \mathbb{N}$. From Lebesgue's theorem we have $$0 = \int_0^1 f_m(x) f_{n_k}(x)\,dx \xrightarrow[k \to \infty]{} \int_0^1 f_m(x) f(x)\,dx.$$ Hence $\int\limits_0^1 f_m(x) f(x)\,dx = 0$ for every $m \in \mathbb{N}$, which implies $f(x) = 0$ almost everywhere. Using once more Lebesgue's theorem we get $$1 = \int_0^1 f_{n_k}^2(x)\,dx \xrightarrow[k \to \infty]{} \int_0^1 f^2(x)\,dx = 0.$$ The contradiction proves the statement.