Show that $f_n\to f$ in $L^p$ if and only if $f^p_n \to f^p$ in $L^1.$

Question

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.

Answer 1

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 $p>1$.

Answer 2

The answer will be a consequence of the following results:

Lemma 1: Suppose $f,g\geq0$ are functions in $L_p$ ($p>0$). For $\alpha >1$,

$$ \begin{align} \int|f^\alpha -g^\alpha|^{p/\alpha}\,d\mu\leq \alpha^{p/\alpha}\Big(\int(f+g)^p\,d\mu\Big)^{1-\tfrac1\alpha}\Big(\int|f-g|^p\,d\mu\Big)^{1/\alpha}\tag{1}\label{one} \end{align} $$

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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 $\|f-f_n\|_p\xrightarrow{n\rightarrow\infty}0$.

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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 $$ \begin{align} \int |f^\alpha-g^\alpha|^{p/\alpha}\,d\mu &\leq \alpha^{\frac{p}{\alpha}}\int (f\vee g)^{\frac{\alpha-1}{\alpha}p}|f-g|^{\frac{p}{\alpha}}\,d\mu\\ &\leq \alpha^{\frac{p}{\alpha}}\Big(\int (f\vee g)^p\,d\mu\Big)^{1-\frac{1}{\alpha}}\Big(\int |f-g|^p\,d\mu\Big)^{\frac{1}{\alpha}}\\ & \leq \alpha^{\frac{p}{\alpha}}\Big(\int (f+g)^p\,d\mu\Big)^{1-\frac{1}{\alpha}}\Big(\int |f-g|^p\,d\mu\Big)^{\frac{1}{\alpha}} \end{align} $$

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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 $$ \int|f^\beta-g^\beta|^{q/\beta}\,d\mu \leq \int|f-g|^q\,d\mu $$

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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 $\|f^p-f^p_n\|_1\xrightarrow{n\rightarrow\infty}0$.

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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 $$ \int|f^\beta -g^\beta|^{q/\beta}\leq \int|f-g|^p\,d\mu$$

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Comments:

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

2. 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.