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Possible Duplicate:
$\lim_{n\rightarrow \infty}\int_0^1f_nhdm=\int_0^1fhdm$, prove $f\in L^p(m)$ , where $1\le p<\infty$.

Can anyone help with this question?

When ${f_n}$ is defined on [0,1], $ ||f_n||_p\le1$, $1, $f$ is integrable on $[0,1]$, $m$ is Lebesgue measure, and $\lim_{n\rightarrow \infty}\int_0^1f_ng \, dm=\int_0^1fg \, dm,$ for any $g\in L^{\infty}(m)$.

Prove $f\in L_p$

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    This (essentially unanswered) question is exactly the same -- except that the case $p =1$ is allowed in the other question: http://math.stackexchange.com/q/241076/2012-11-21

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Since $L^\infty[0,1]$ is dense in $L^q[0,1]$, $ \|f\|_p=\sup\{\int_0^1fg\,dm: \|g\|_q=1\}=\sup\{\int_0^1fg\,dm: g\in L^\infty[0,1]\mbox{ and }\|g\|_q=1\}. $ Note that, for $g\in L^\infty$ with $\|g\|_q=1$, $ \left|\int_0^1fg\,dm\right|=\lim\,\left|\int_0^1f_ng\,dm\right|\leq \|f_n\|_p\,\|g\|_q\leq1. $ So $\|f\|_p\leq1$.