If $f_n \rightarrow f$ in $L^p$, Does $f_n^p \rightarrow f^p$ in $L^1$?

Question

If $f_n \rightarrow f$ in $L^p$, Does $f_n^p \rightarrow f^p$ in $L^1$?

This seems a very trivial fact, but just from looking at the norm definition of convergence, it does not seem entirely clear how $||f_n - f||_p \rightarrow 0$ would imply $||f_n^p - f^p||_1 \rightarrow 0$.

Does the statement hold?

Answer 1

I will assume that $ 1<p<\infty$. Let $q$ be the conjugate index. $|x^{p}-y^{p} |\leq p(|x|^{p-1}+|y|^{p-1}) |x-y|$ by MVT. To show that $\int |f|^{p-1} |f-f_n| \to 0$ apply Holder's inequality: $\int |f|^{p-1} |f-f_n| \leq \|f-f_n\|_p (\int |f|^{(p-1)q})^{1/q}$. Since $q(p-1)=p$ we see that $\int |f|^{p-1} |f-f_n| \to 0$. Similarly, $\int |f_n|^{p-1} |f-f_n| \to 0$.