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Let $(X,m)$ be a measure space, $(f_n)_n, (g_n)_n, h \in L^2(m)$. I would like to prove that if $0 \leq f_n \leq g_n$, $g_n \rightarrow h$ in $L^2$ and $\int f_n^2 \rightarrow \int h^2$ then $f_n \rightarrow h$ in $L^2$.

My try:

$$ \int (f_n - h)^2 = \int f_n^2 - 2f_nh + h^2, $$ and since $\int f_n^2 \rightarrow \int h^2 $ I am done if I can prove $$ \int f_n h \rightarrow \int h^2. $$ I would be tempted to write $$ \left|\int (f_n - h) h \right| \leq \left| \int (g_n - h)h \right| \rightarrow 0, $$ but I only know that $f_n \leq g_n$.

Aspera
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2 Answers2

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As it is stated now with John's correction the thesis comes from my other question, which applied to this case shows that $$\sqrt{f_n^2} = f_n \rightarrow h = \sqrt{h^2} \text{ in } L^2 \quad \Leftrightarrow \quad f_n^2 \rightarrow h^2 \text{ in } L^1. $$ But the latter is equivalent to requiring $\int f_n^2 -h^2 \rightarrow 0, $ which is one of the hypotheses. Thus reading backwards this answer we have the thesis.

Community
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Aspera
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First note that $$ g_n^2=(g_n-f_n)^2+2f_n(g_n-f_n)+f_n^2. $$ Hence $$ \int g_n^2-\int f_n^2=\int (g_n-f_n)^2+2\int f_n(g_n-f_n)\ge\int (g_n-f_n)^2=\|g_n-f_n\|^2\ge 0. $$ And as $\int g_n^2\to\int h^2$ and $\int f_n^2\to\int h^2$, then $\|g_n-f_n\|\to 0$. But $$ \|h-f_n\|\le \|h-g_n\|+\|g_n-f_n\|\to 0, $$ and hence $\|h-f_n\|\to 0$.

Yiorgos S. Smyrlis
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