Let $f_n\in L^2([0,1])$ and $\lvert\lvert f_n\rvert\rvert=1$ $\forall n$ how can we show that there exists $f\in L^2([0,1])$ and a subsequence $f_{n_k}$ which converges to f in $L^2([0,1])$
if i can show this then i can do the rest. but this part is necessary
As $\textbf {Hint}$: $L^2([0,1])$ has a countable dense subset.
We cannot in general extract a subsequence which converges for the $\mathbb L^2$ norm: take $f_n(x):=e^{2i\pi nx}$: we have boundedness property, and if it converges in $\mathbb L^2$ to an $f$, then $\int_0^1 f(x)e^{2i\pi kx}\mathrm dx=0$ for each $k$, hence $f=0$. But $\lVert f_n\rVert_2=1$.
However, it is possible to extract a subsequence which converges weakly, that is, there is an $f$ such that for any $g\in\mathbb L^2(0,1)$, $$\int_0^1 f_n(x)g(x)\mathrm dx\to \int_0^1 f(x)g(x)\mathrm dx,\quad n\to \infty.$$ Indeed, we use the hint: take $(h_N)_{N\geqslant 1}$ a dense sequence in $\mathbb L^2(0,1)$. By a diagonal argument and using boundedness, there is an increasing sequence $n_k\uparrow\infty$ such that for each $N$, $$\int_0^1 f_{n_k}(x)g_N(x)\mathrm dx\to \int_0^1 f(x)g_N(x)\mathrm dx,\quad k\to \infty.$$ Then we prove for an arbitrary $g$ by an approximation argument.