Is there a sequence $(f_n)\in\ L^2([0,1])$, s.t. $\lVert f_n\rVert_2=1$, $\forall n$, but it has no convergent subsequences in $L^2([0,1])$ ?

Question

Is there a sequence $(f_n)\in\ L^2([0,1])$, s.t. $\lVert f_n\rVert_2=1$, $\forall n$, but it has no convergent subsequences in $L^2([0,1])$ ?

We know at least $(f_n)$, is not convergent in the normal sense$ (because\ L^2convergence\Rightarrow a.e.convergence) $

Is it a ''famous'' function, or can we construct it (I think we must sin, cos or characteristic function bring into play) ?

Answer 1

Any sequence of mutually orthogonal functions of unit norm has this property, because $\|f_m-f_n\| = \sqrt{2}$ for all $m\ne n$. For example, $f_n(t)=\exp(2\pi i n t)$.