Prove that $\int \vert \ \vert f_n\vert-\vert f_n-f\vert^p -\vert f\ \vert \ \vert d\mu \to 0$, if $\sup_n\vert f_n \vert\leq C$

measure-theory

Let $f_n\in L(\mu)$, $1\leq p<\infty$ be such that $\exists C>0$ such that $\sup_n \vert f_n \vert^p\leq C$ and also $f_n\to f$. So prove that if $n\to\infty$, then $$\int \vert \ \vert f_n\vert-\vert f_n-f\vert^p -\vert f\ \vert \ \vert d\mu \to 0$$

Since the integrand is a positive functions, I tried to use Fatou's Lema for prooving something like $\lim \sup \int \cdots d\mu =0$, but wasn't successeful.

Searching I little more, I found something similar on "Lieb and Loss – Analysis" pg.21
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I could understand (or almost could) this proof, but I found it too "tricky" and maybe there is another diffrent way…

But also about this proof… I couldn't link the hypothesis of this theorem with my question, is there something missing? If $\mu(X)<\infty$ I could do $\vert f_n \vert^p \leq \sup \vert f_n \vert ^p \leq C \implies \int \vert f_n \vert ^pd\mu < C\mu(X)<\infty$ … otherwise I couldn't.

There are some steps of this proof that I couldn't get too, if anyone could give me hints of how to proceed by another way I'd be grateful!

Best Answer

What you are claiming:

Let $f_n\in L(\mu)$, $1\leq p<\infty$ be such that $\exists C>0$ such that $\sup_n \vert f_n \vert^p\leq C$ and also $f_n\to f$. So prove that if $n\to\infty$, then $$\int \vert \ \vert f_n\vert-\vert f_n-f\vert^p -\vert f\ \vert \ \vert d\mu \to 0$$

is in fact not true!

For a simple counter-example, take Lebesgue measure on $\mathbb R$, and consider $f_n(x) = c \, 1_{[n,n+1]}(x)$ for some $c>0$. Then $f_n \to f(x) := 0$, and $f_n$ is uniformly bounded in the sup-norm. Furthermore, $$ \int_\mathbb{R} ||f_n|-|f| - |f_n-f|^p|= \int_\mathbb{R} |c - c^p| \, 1_{[n,n+1]} = |c-c^p|>0, $$ if $p\ne 1$ and $c\ne 1$...

Maybe you made a mistake when copying this question?

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