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I'm trying to solve an exercise, i.e.,

Construct a sequence $\left(f_n\right)$ in $L^1(0,1), f_n \geq 0$, such that:

  • (i) $f_n \rightarrow f$ weakly $\sigma\left(L^1, L^{\infty}\right)$,
  • (ii) $\left\|f_n\right\|_1 \rightarrow\|f\|_1$,
  • (iii) $\left\|f_n-f\right\|_1 \nrightarrow 0$.
  1. Could you confirm if my below attempt is fine?
  2. Is there a more intuitive example of such $(f_n, f)$?

Let $T:=1$. Let $f:\mathbb R \to \mathbb R$ be $T$-periodic such that $f(x)= 0$ if $x \in [0, 1/2)$ and $f(x)=2$ if $x \in [1/2, 1)$. Let $\bar f := \frac{1}{T} \int_0^T f (t) \, dt.$ Then $\bar f=1$. We define $f_n:(0, 1) \to \mathbb R$ by $f_n (x) := f(nx)$ for all $x \in (0, 1)$. We have

  • (1) $f_n \to \bar f$ in the weak topology $\sigma (L^1 (0, 1), L^\infty (0, 1))$, and
  • (2) $\lim_n \|f_n-\bar f\|_{L^1(0, 1)} = \frac{1}{T} \int_0^{T} |f-\bar f|$.

On the other hand, $\frac{1}{T} \int_0^{T} |f-\bar f| = 1>0$. By symmetry, $\|f_n\|_1 = \|f\|_1 = 1$ for all $n$.

Analyst
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  • The example looks fine. The weak convergence in $L^1$ holds because the sequence is bounded, e.g., in $L^2$ and the measure of the space $[0,1]$ is finite (hence the sequence is equiintegrable). I don’t think there is a better example, your sequence has to be equiintegrable (cannot concentrate too much) and the measure space has finite measure, so the only thing I can imagine is the sequence has to oscillate – Lorenzo Pompili Apr 08 '23 at 07:35

1 Answers1

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Notice that many examples of this type can be obtained from a previous posting of the OP, which can be seen as a Corollary to Fejer's lemma.

Let $f\in L_1[0,1]\setminus\operatorname{span}(\mathbb{1})$. Extend $f$ to $\mathbb{R}$ by $1$-periodization so that $f(x+1)=f(x)$ for all $x\in\mathbb{R}$. Define $f_n(x)=f(nx)$. Clearly $\|f\|_{L_1[0,1]}=\|f_n\|_{L_1[0,1]}$, and by Fejer's lemma, for any $g\in L_\infty[0,1]$ $$\int^1_0 gf_n=\int^1_0g(x)f(nx)\,dx\xrightarrow{n\rightarrow\infty}\Big(\int^1_0 f\Big)\Big(\int^1_0 g\Big)$$ that is $f_n\xrightarrow{n\rightarrow\infty}\int f$ in the weak topology $\sigma(L_1[0,1],L_\infty[0,1])$. Let $\overline{f}:=\int^1_0f$. Since $f$ is $1$-periodic and not constant, $$\int^1_0 |f_n-\overline{f}|=\int^1_0|f(nx)-\overline{f}|\,dx=\int^1_0|f(u)-\overline{f}|\,du>0$$

Mittens
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