Consider $\Omega$ an open non-empty set of $\mathbb{R}^n$; We know that $(L^p(\Omega),\left\lVert \cdot\right\rVert_{L^p})$ is an infinite dimensional vector space, thus thanks to Riesz's lemma we know that the closed unit ball $B_{L^p}$ is not compact (with respect to $\left\lVert \cdot\right\rVert_{L^p}$).
Let $\Omega=(0,1)$: I'm wondering if $B_{L^2((0,1))}$ is relatively compact in $(L^1((0,1),\left\lVert \cdot\right\rVert_{L^1}$).
I'm asking directly the "relactive compactness" because, if I'm not wrong, $B_{L^2((0,1))}$ is not closed in $(L^1((0,1)),\left\lVert \cdot\right\rVert_{L^1})$.
I know, thanks to a corollary of Riesz-Frechet-Kolmogorov theorem, that if $B_{L^2((0,1))}$ is bounded, and it holds that, for every $f\in B_{L^2((0,1))}$, $$\lim_{v\rightarrow 0} \left\lVert \tau_v\hat{f}-\hat{f}\right\rVert_{L^1}=0,$$ where $\hat{f}$is the extension of $f$ to the whole $\mathbb{R}$ ($0$ outisde $(0,1)$), $v\in\mathbb{R}$ and $\tau_v(f)(x)=f(x+v)$, then $B_{L^2((0,1))}$ is relatively compact in $(L^1((0,1),\left\lVert \cdot\right\rVert_{L^1})$.
I studied this theorem, but apart from remembering the claim I have no idea how to check these properties: I have the feeling this set is at least bounded, but again, no idea on how to prove it. This whole question is probably nonsense since it was my thought and not an exercise.
Any hint or solution would be much appreciated... Thanks in advance.
The $f_n: t \longmapsto e^{2i\pi nt}$ are in $B_{L^2}$, and converge weakly as $L^1$ functions to zero: however, their $L^1$ norm is $1$ so there isn’t any convergent subsequence.