Show that $f_n\to f$ in $L^p$ if and only if $f^p_n \to f^p$ in $L^1.$ Let $f_n$ be a sequence of nonnegative measurable functions in $L^p(\mathbb R)$ for some $1<p<\infty.$ Show that $f_n\to f$ in $L^p$ if and only if $f^p_n \to f^p$ in $L^1.$
I showed that if $f,f_n\in L_p$ and $f_n \to f$ pointwise a.e., then $||f−f_n||_p \to 0$ iff $||f_n||_p \to||f||_p$. But I found that $||f_n||_p \to||f||_p$ was different from $f^p_n \to f^p$ in $L^1.$ So, it didn't help with my problem. I got stuck on this problem.
 A: If $f_n\to f$ in $L^p(\mathbb{R})$ then with the convexity of $x^p$ for $p>1$, $$|x^p-y^p|\le p(x+y)^{p-1}|x-y|$$ Therefore by Holder's inequality with $\frac{1}{q}+\frac{1}{p}=1$, $$\therefore \int|f^p-f_n^p|\le p\|(f+f_n)^{p-1}\|_q\|f-f_n\|_p$$ But $(p-1)q=p$, $$\therefore \|f^p-f_n^p\|_1\le p\|(f+f_n)^p\|_1^{1/q}\|f-f_n\|_p\le c\|f-f_n\|_p\to0$$
If $f_n^p\to f^p$ in $L^1(\mathbb{R})$ (with $f_n,f\ge0$), then $$\left|f-f_n\right|^p\le\left||f|^p-|f_n|^p\right|$$ $$\therefore \int|f-f_n|^p\le\int|f^p-f_n^p|\to0$$ The first inequality uses $|x-y|^p\le|x^p-y^p|$ valid for $p>1$.
A: The answer will be a consequence of the following results:
Lemma 1: Suppose $f,g\geq0$ are functions in $L_p$ ($p>0$). For $\alpha >1$,
$$
\begin{align}
\int|f^\alpha -g^\alpha|^{p/\alpha}\,d\mu\leq \alpha^{p/\alpha}\Big(\int(f+g)^p\,d\mu\Big)^{1-\tfrac1\alpha}\Big(\int|f-g|^p\,d\mu\Big)^{1/\alpha}\tag{1}\label{one}
\end{align}
$$

With $\alpha=p$, $p>1$,  we get one part of the problem:
$$\|f^p-f^p_n\|_1\leq p\, \|f+f_n\|^{p/(1-p)}_p\|f-f_n\|_p\xrightarrow{n\rightarrow\infty}0$$
when $\|f-f_n\|_p\xrightarrow{n\rightarrow\infty}0$.

Proof of Lemma 1
Using the mean value theorem we get that
$$
|f^\alpha - g^\alpha|\leq \alpha(f\vee g)^{\alpha-1}|f-g|
$$
where $f\vee g=\max(f,g)$. From this, it follows that
$$
\begin{align}
 \int |f^\alpha-g^\alpha|^{p/\alpha}\,d\mu &\leq \alpha^{\frac{p}{\alpha}}\int (f\vee g)^{\frac{\alpha-1}{\alpha}p}|f-g|^{\frac{p}{\alpha}}\,d\mu\\
   &\leq \alpha^{\frac{p}{\alpha}}\Big(\int (f\vee g)^p\,d\mu\Big)^{1-\frac{1}{\alpha}}\Big(\int |f-g|^p\,d\mu\Big)^{\frac{1}{\alpha}}\\
& \leq \alpha^{\frac{p}{\alpha}}\Big(\int (f+g)^p\,d\mu\Big)^{1-\frac{1}{\alpha}}\Big(\int |f-g|^p\,d\mu\Big)^{\frac{1}{\alpha}}
\end{align}
$$

The other direction is simpler and can be obtain in a similar way:
Lemma 2:  Suppose $f,g\in L_q$( $q>0$), and $f,g\geq0$. If  $0<\beta<1$, then
$$
  \int|f^\beta-g^\beta|^{q/\beta}\,d\mu \leq \int|f-g|^q\,d\mu
$$

With $F=f^p,G=g^p\in L_1$, $\beta=1/p$ and $q=1$ we get
$$
\begin{align}
\int|f-g|^p\,d\mu\leq \int |f^p-g^p|\,d\mu\tag{2}\label{two}
\end{align}
$$
which can be use to solve the other part of problem:
$$\|f-f_n\|^p_p\leq \|f^p-f^p_n\|_1\xrightarrow{n\rightarrow\infty}0$$
when $\|f^p-f^p_n\|_1\xrightarrow{n\rightarrow\infty}0$.

Proof of Lemma 2
First, recall that for $0<\beta<1$,  inequality $1+x^\beta<(1+x)^\beta$ for all $x\geq0$. (This can be shown by looking at the auxiliary function $\phi(x)=(1+x)^\beta -x^\beta$ and analyzing it monotone behavior in $[0,\infty)$.
From this, it follows that
that
$$
 (a+b)^\beta\leq a^\beta + b^\beta,\qquad a,b\geq0
$$
Thus,
$$a^\beta = (a-b+b)^\beta\leq (|a-b|+b)^\beta\leq |a-b|^\beta +b^\beta$$
reversing the roles of $a$ and $b$ leads to
$$|a^\beta-b^\beta|\leq |a-b|^\beta$$
Putting things together we obtain that
$$
\int|f^\beta -g^\beta|^{q/\beta}\leq \int|f-g|^p\,d\mu$$

Comments:

*

*The second part of the proof has been updated to use a method consistent with the strategy of the first part.


*This is not only to produce solutions to the OP,  but also the emphasize the inequalities $\eqref{one}$ and $\eqref{two}$ which are very useful.
