Convergences of $ f_n \in L^p$

Question

I'm trying to solve the following task:

Let $(\Omega,\mathfrak{A},\mu)$ be a measure space, and $f,f_1,f_2,\dots \in L^P(\Omega,\mathfrak{A},\mu)$ with $p\in [1,\infty[$. Ssuppose that, as $n\rightarrow \infty$, $f_n(\omega)\rightarrow f(\omega)$ for $\mu-$a.e. $\omega \in \Omega$ and that $||f_n||_p\rightarrow ||f||_p$.

Let sign: $\mathbb{R}\rightarrow \{-1,1\}$ denote a function such that $|x|=($sign $x$)$x$ for all $x \in \mathbb{R}$, and write $$f_n^*:=f_n\mathbb{1}_{\{|f_n|\leq|f|\}} + (sign\ f_n)|f|\mathbb{1}_{\{|f_n|>|f|\}}$$ for every $n \in \mathbb{N}$

1. Show that $||f_n^* - f||_p\rightarrow 0$ as $n\rightarrow \infty$
2. Show that $||f_n - f_n^*||_p\rightarrow 0$ as $n\rightarrow \infty$ Conclude that $f_n \rightarrow f \in L^p(\Omega,\mathfrak{A},\mu)$

My first thought was that I could use dominated convergence theorem to show $$\lim\limits_{n\rightarrow \infty}\left(\int |f_n^* -f|^p d\mu\right)^{\frac{1}{p}}$$ Since $f_n \in L^p(\Omega,\mathfrak{A},\mu) $ but I'm not quiet exactly sure how to find the dominated function.

My second thought was to use the Minkowski inequality:

$$\lim\limits_{n\rightarrow \infty}||f_n^* -f||_p = \lim\limits_{n\rightarrow \infty}||f_n\mathbb{1}_{\{|f_n|\leq|f|\}} + (sign\ f_n)|f|\mathbb{1}_{\{|f_n|>|f|\}} -f||_p \leq $$

$$\lim\limits_{n\rightarrow \infty} ||f_n\mathbb{1}_{\{|f_n| \leq|f|\}}||_p + ||(sign\ f_n)|f|\mathbb{1}_{\{|f_n|>|f|\}} -f ||_p $$ $$=||f\mathbb{1}_{\{|f_n| \leq|f|\}}||_p + \lim\limits_{n\rightarrow \infty} ||(sign\ f_n)|f|\mathbb{1}_{\{|f_n|>|f|\}} -f ||_p$$

But I don't know any further or which is the right way.

Answer 1

We have defined $$f_n^*:=f_n\mathbb{1}_{\{|f_n|\leq|f|\}} + (sign\ f_n)|f|\mathbb{1}_{\{|f_n|>|f|\}}$$ for every $n \in \mathbb{N}$.

Since as $n\rightarrow \infty$, $f_n(\omega)\rightarrow f(\omega)$ for $\mu-$a.e. $\omega \in \Omega$, we have that, $$\mathbb{1}_{\{|f_n|\leq|f|\}} \rightarrow \mathbb{1}_{\Omega} \:\:\: \mu\textrm{-a.e.}$$ and $$\mathbb{1}_{\{|f_n|>|f|\}} \rightarrow 0 \:\:\: \mu\textrm{-a.e.}$$ So $$f_n^*(\omega)\rightarrow f(\omega)\:\:\: \mu\textrm{-a.e.}$$ Now, since $ |f_n|\mathbb{1}_{\{|f_n|\leq|f|\}} \leq |f|\mathbb{1}_{\{|f_n|\leq|f|\}}$, we have $$|f_n^*|\leq |f_n|\mathbb{1}_{\{|f_n|\leq|f|\}} + |f|\mathbb{1}_{\{|f_n|>|f|\}}\leq|f|\mathbb{1}_{\{|f_n|\leq|f|\}} + |f|\mathbb{1}_{\{|f_n|>|f|\}}=|f|$$ So we can apply de Dominated Convergence Theorem (for $L^P$) and we get $f_n^* \rightarrow f \in L^p(\Omega,\mathfrak{A},\mu)$, which means $$||f_n^* - f||_p\rightarrow 0 \textrm{ as } n\rightarrow \infty$$ To complete the proof, note that $$|f_n-f_n^*|^p= (2\max(|f_n|,|f_n^*|))^p=2^p\max(|f_n|^p,|f_n^*|^p)\leq 2^p(|f_n|^p +|f_n^*|^p)$$ So, consider $$h_n = 2^p(|f_n|^p +|f_n^*|^p) - |f_n-f_n^*|^p$$ for every $n \in \mathbb{N}$. We have that, for every $n \in \mathbb{N}$, $h_n\geq 0$. Since $f_n \rightarrow f$ $\mu$-a.e. and $f_n^* \rightarrow f$ $\mu$-a.e., we have that $h_n \rightarrow 2^{p+1}|f|^p$ $\mu$-a.e.. Using Fatou's Lemma, we get \begin{align} \int 2^{p+1}|f|^p d\mu &=\int \lim\inf h_n d\mu \leq \lim\inf \int h_n d\mu = \\&= \lim\inf \left (\int 2^p|f_n|^p d\mu +\int 2^p|f_n^*|^p d\mu -\int |f_n-f_n^*|^p d\mu\right)= \\ & = \int 2^p|f|^p d\mu +\int 2^p|f|^p d\mu -\lim\sup\int |f_n-f_n^*|^p d\mu \end{align} So we have $$ \int 2^{p+1}|f|^p d\mu \leq \int 2^{p+1}|f|^p d\mu -\lim\sup\int |f_n-f_n^*|^p d\mu$$ Since $f\in L^P(\Omega,\mathfrak{A},\mu)$, $\int |f|^p d\mu<+\infty$, and we get $$ \lim\sup\int |f_n-f_n^*|^p d\mu\leq 0$$ So we have that $||f_n - f_n^*||_p\rightarrow 0$ as $n\rightarrow \infty$.