# Convergence in $L^p$ of $f_n$ implies convergence in $L^1$ of $|f_n|^p$ and $f_n^p$

I have the following question about $L^p$ spaces:

Suppose that $f,f_1,f_2,\ldots$ are functions in $L^p$ for some $p \geq 1$ and the sequence converges in $L^p$ to $f$, i.e. $||f_n-f||_p \to 0$.

Does this imply that the sequence $|f_1|^p,|f_2|^p,\ldots$ converges in $L^1$ to $|f|^p$?

Is it also true if we remove the absolute value? That is, the sequence $f_1^p,f_2^p,...$ converges to $f^p$, in case $p$ doesn't create problem with power (for example $p=3$).

I am using necessary and sufficient condition from the Vitali's convergence theorem and it seems to work.

Do you have some advices or some references for this result?

Many thanks

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What are you asking? – checkmath Mar 30 '12 at 15:12
Is true that $||\,|f_n|^p-|f|^p \, ||_1 \to 0$? – Kolmo Mar 30 '12 at 15:19
Hint you have $\int |f_n|^p \to \int |f|^p$. – checkmath Mar 30 '12 at 15:19
So if you have $|f_n|$ dominated by $|f|$ then it follows directly! – checkmath Mar 30 '12 at 15:26
@chessmath you mean $|f_n| \leq |f|$ a.e for all n, then $|| \, |f_n|^p -|f|^p \, || = \int |f|^p-|f_n|^p$ and it is ok. But I dont have that dominance. – Kolmo Mar 30 '12 at 15:33

If $a,b\in \mathbb{R}$ and $p\ge1$, then it follows from the mean value theorem that $$|a^p-b^p|\le p\max(|a|,|b|)^{p-1}|a-b|\le p(|a|+|b|)^{p-1}|a-b|.$$ Let $q$ be the conjugate exponent of $p$. Then $$\int\bigl|\,|f_n|^p-|f|^p\,\bigr|\le p\int\bigl(|f_n|+|f|\bigr)^{p-1}|f_n-f|\le p\Bigl(\int\bigl(|f_n|+|f|\bigr)^{(p-1)q}\Bigr)^{1/q}\Bigl(\int|f_n-f|^{p}\Bigr)^{1/p},$$ that is $$\||f_n|^p-|f|^p\|_1\le p(\||f_n|+|f|\|_p)^{p/q}\|f_n-f\|_p\le p(\|f_n\|_p+\|f\|_p)^{p/q}\|f_n-f\|_p,$$ proving that $|f_n|^p\to|f|^p$ in $L^1$.

Observe that we have proved that the operators $T_1, T_2\colon L^p\to L^1$ defined by $T_1(f)=|f|^p$ and $T_2(f)=f^p$( when $T_2$ makes sense) are locally Lipschitz.

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Thanks very much Julian. So direct and clean. Thanks again. – Kolmo Mar 31 '12 at 9:13

If we assume that the result is not true, we would be able to find a subsequence $\{f_{n_k}\}$ and a $\delta>0$ such that $\lVert |f_{n_k}|^p-|f|^p\rVert_{L^1}\geq \delta_0$. Taking if necessary a further subsequence, we can assume that $f_{n_k}$ converges to $f$ almost everywhere. We denote $g_k:=|f_{n_k}|^p$, and let $h_k:=|g_k|+|f|^p-|g_k-|f|^p|$. Then almost everywhere, $h_k$ converges to $2|f|^p$, and since $h_k\geq 0$, by Fatou lemma: $$2\lVert f\rVert_{L^p}^p\leq \liminf_{k\to +\infty}\int (|g_k|-|g_k-|f|^p|)+\int|f|^p$$ and since $\int g_k\to \int |f|^p$ we have that $\limsup_{k\to \infty}\int |g_k-|f||^p=\limsup_{k\to \infty}\int ||f_{n_{k}}|^p-|f|^p|=0$. I followed a method similar to here.

For $f_n^p$ just choose $g_k =f_{n_k}^p$.

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If I take a=2, b=1 and p=2 we have $|2^2-1^2|=3>1=|2-1|^2$ – Kolmo Mar 30 '12 at 15:48
@Kolmo My attempt was wrong, I hope this one is correct. – Davide Giraudo Mar 30 '12 at 16:17
thanks for the responce. I have a doubt. This reasoning shows that $\limsup_{k\to\infty} \int |\ |f_{n_k}|^p-|f|^p \ | \leq 0$. Why this should be true for the whole sequecence? – Kolmo Mar 31 '12 at 9:02
Let $a_n:=\int ||f_n|^p-|f_n||$. We have shown that for each subsequence of $\{a_n\}$ we can find a subsequence which converges to $0$. – Davide Giraudo Mar 31 '12 at 10:53
so $a_{n_k}=\int ||f_{n_k}|^p-|f|^p|$. Since $f_n$ converge in $L^p$, the same does $f_{n_k}$ and so we know that exists a subsequence of $f_{n_k}$ that converge a.e. to f. Then if we say $h_l=|f_{n_{k_l}}|^p$ this converges a.e. to $|f|^p$ and $||h_l||_1 \to |||f|^p||_1$. So by the trick you linked $a_l \to 0$. Did I understand correct. I guess yes because it works. Thanks. – Kolmo Mar 31 '12 at 11:14