$L^1(\mathbb R)$ convergence of min and max functions.

Question

Let $f_k , f \in L^1(\mathbb R)$ , with $f_k,f \geq 0$ a.e. in $(0,1)$. Suppose that $f_k\rightarrow f$ pointwise a.e. in $(0,1)$ and that:

$\int_0^1 f_n dx \rightarrow \int_0^1 fdx$

Prove that:

1) $\max(f_n,f)\rightarrow f$ in $L^1(\mathbb R)$

2) $\min(f_n,f)\rightarrow f$ in $L^1(\mathbb R)$

3) $f_n \rightarrow f $ in $L^1(\mathbb R)$

First I rewrite min and max functions as:

$\max{(f_n,f)}=\frac{f_n+f}{2}+\frac{|f_n-f|}{2}$

$\min{(f_n,f)}=\frac{f_n+f}{2}-\frac{|f_n-f|}{2}$

Now, does the pointwise convergence a.e. and $\int_0^1 f_n dx \rightarrow \int_0^1 fdx$ imply:

$\lim_{n \to \infty}\int_0^1 f_n dx= \int_0^1 f dx =\int_0^1 \lim_{k \to \infty} f_{n_k} dx$ ?

With this statement being true, it is simple to obtain the results of the proof, but I'm not sure whether it is. Could You please give me some more ideas if mine is wrong.

Thanks

Answer 1

$\int_0^{1} ((f-f_n)^{+})\, dx \to 0$ by DCT (where $f^{+}(x)=\max \{f(x),0\}$). (Note that $0 \leq (f-f_n)^{+} \leq f$). Since $\int_0^{1} f_n(x) \, dx \to \int_0^{1} f(x) \, dx$ we get $\int |f_n(x)-f(x)|\, dx =2\int_0^{1} (f-f_n)^{+}\, dx -\int_0^{1} (f_n(x)-f(x)) \, dx \to 0$. Also $|\max \{f_n,f\}-f\} \leq |f_n-f|$ and $|\min \{f_n,f\}-f\}\leq |f_n-f|$

Answer 2

I have to admit that I do not understand the point of the last displayed equation (the first equality is by assumption and the second one is also true because the pointwise limit of a subsequence is also $f$). The key point here is that we can use your expression for the $\min$ and $\max$ combined with the convergence $$ \lim_{n\to +\infty}\int_0^1\left\lvert f_n(x)-f(x)\right\rvert\mathrm dx=0, $$ which can be established by applying Fatou's lemma to the sequence $\left(g_n\right)_{n\geqslant 1}$, where $$ g_n(x)=f_n+f-\left\lvert f_n(x)-f(x)\right\rvert. $$