# Unit ball in $L_2[0,1]$ is not relatively compact in $L_1[0,1]$

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.

• Pick the family of functions $f_n(x)=\sqrt{n}1_{[0,1/n]}$. Then check what happens if you pick $n>2/\delta$ and $h=\delta/2$. – Severin Schraven Nov 12 '20 at 20:36
• Then we obtain that $\left(\int_0^1|f_n(x+h) - f_n(x)|dx\right) = \frac{1}{\sqrt{n}}$. So I don't see what can we say next – Mikhail Goltvanitsa Nov 12 '20 at 21:17
• If you also use the correct exponents ($p=2$), then you will get $\sqrt{2}$. – Severin Schraven Nov 12 '20 at 21:56
• But we apply criterion for p=1, so the correct exponent is 1. Not so? – Mikhail Goltvanitsa Nov 13 '20 at 5:22
• No, we don't apply the criterion for $p=1$. The space is $L_2$ (so $p=2$). Why should we apply it for $p=1$? (Note that my trick works for any $p\in [1,\infty)$, we would then consider $n^{1/p} 1_{[0,1/n]}$, the point is that the $p$-norm is equal to $1$ and the support gets very small, such that the support of the shifted function is disjoint from the unshifted support). – Severin Schraven Nov 13 '20 at 9:29

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.