I want to prove that the following set $$ B = \{f\in L_2[0,1]: \int_{0}^{1}|f|^2d\mu\leq 1\} $$ is not relatively compact in $L_1[0,1]$.
I know the general ctiterion of relatively compactness in $L_p[a,b]$. A set $T \subset L_p[a,b]$ is relatively compact iff it is bounded and for every $\varepsilon > 0$ there exists $\delta > 0$ such that for every $h$ with the property $|h| < \delta$ we have $$ \left(\int_a^b|f(x+h) - f(x)|^pdx\right)^{\frac{1}{p}} < \varepsilon $$
I don't know how to use this criterion here. Maybe there are some other approaches.
The unit ball in $L^2([0,1])$ is not relatively compact in $L^p([0,1])$ for any $p\in [1,2]$. Fix $p\in [1,2]$ and define for $n\in \mathbb{N}$ the set $A_n = \bigcup_{j=0}^{n-1} [2j/(2n), (2j+1)/(2n)]$ (i.e. we divide the unit interval into $2n$ pieces of equal length and kick out every second one) and the function $f_n = 1_{A_n}$. Then we have $$ \int_0^1 \vert f_n(x)\vert^2 dx = \int_0^1 f_n(x) dx = 1/2 \leq 1. $$ On the other hand we have for every $n\in \mathbb{N}$ and every $p\in [1,2]$ $$ \int_0^1 \vert f_n(x+1/(2n)) - f_n(x) \vert^p dx = 1. $$ Thus, for $\varepsilon = 2/3$ the criterion fails and you arrive at the desired conclusion.