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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.

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  • $\begingroup$ 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 $h=\delta/2$. $\endgroup$ – Severin Schraven Nov 12 '20 at 20:36
  • $\begingroup$ 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 $\endgroup$ – Mikhail Goltvanitsa Nov 12 '20 at 21:17
  • $\begingroup$ If you also use the correct exponents ($p=2$), then you will get $\sqrt{2}$. $\endgroup$ – Severin Schraven Nov 12 '20 at 21:56
  • $\begingroup$ But we apply criterion for p=1, so the correct exponent is 1. Not so? $\endgroup$ – Mikhail Goltvanitsa Nov 13 '20 at 5:22
  • $\begingroup$ 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). $\endgroup$ – Severin Schraven Nov 13 '20 at 9:29
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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.

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  • $\begingroup$ Thank you, very much! I think that this is a useful trick. $\endgroup$ – Mikhail Goltvanitsa Nov 15 '20 at 17:29
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    $\begingroup$ Indeed, usually there are three tricks one can try. Oscillation (what I used here), concentration of mass (what I falsely tried to use in the comments above, sorry for that) or loss of mass (if you have an unbounded domain, shifting "to infinity"). Those three classical tricks allow for weakly converging sequence that do not converge strongly (respectively do not have any strong converging subsequence). $\endgroup$ – Severin Schraven Nov 15 '20 at 17:42

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