My question is about this:
Convergence in $L^p$ of $f_n$ implies convergence in $L^1$ of $|f_n|^p$ and $f_n^p$
It was shown that the author's question was indeed true by the use of MVT.
Is the converse also true?
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No, the converse is not true.
Example Let
$$f_n(x) := \begin{cases} -1 & x \in \left[0,\frac{1}{2} \right] \\ 1 & x \in \bigg(\frac{1}{2},1\bigg]. \end{cases}$$
Then
$$\int_0^1 ||f_n(x)|-1| \, dx = 0$$
i.e. $|f_n| \to 1$ in $L^1([0,1])$. On the other hand, $$\int_0^1 |f_n(x)-1| \, dx = 2 \int_0^{\frac{1}{2}}1 \, dx = 1.$$ This means that $f_n \not \to 1$ in $L^1([0,1])$.
Similary, one can show that $f_n^2 \to 1$ in $L^1([0,1])$ but, again, $f_n$ does not converge to $1$ in $L^2([0,1])$.
saz
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