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This is one of the past qual question. Suppose $\phi$ is a real valued measurable function on $\mathbb{R}$ such that, for any $f$ in $L^{1} (\mathbb{R})$, the product $f\phi$ is also in $L^{1} (\mathbb{R})$. To prove $\phi$ is essentially bounded.

Seriously, I do not know where to start. I kind of thought approaching the problem by contradiction. It seem I am going nowhere from there.

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See Jonas Meyer's answer to this post. – David Mitra Dec 23 '12 at 0:09
up vote 2 down vote accepted

Define $T:L^1(\mathbb{R})\rightarrow\mathbb{R}$ by $$T(f)=\int_{\mathbb{R}}f\phi$$

Note that $T$ is well defined by hypothesis. Im gonna show that $T$ is a bounded linear function. Indeed, suppose that $f_n\rightarrow f$, hence, we can extract a subsequence of $f_n$ (not relabeled) such that $$f_n\rightarrow f,\ a.e$$ and $$|f_n|\leq g$$

where $g\in L^1$. Therefore we have that $f_n\phi\rightarrow f\phi$ almost everywhere and $|f_n\phi|\leq|g\phi|$ where $g\phi\in L^1$ by hypothesis. Now by using Lebesgue theorem we can conclude that $T(f_n)\rightarrow T(f)$.

Because $T\in (L^1)^\star$, we can find $h\in L^{\infty}$ such that $$T(f)=\int_\mathbb{R}fh,\ \forall\ f\in L^1$$

This implies that $h=\phi$ and hence $\phi\in L^{\infty}$

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Where do you get this dominating function for the $f_n$? – David Mitra Dec 23 '12 at 0:13
Theorem 4.9, page 94 from Brezis book: Brezis, Haim Functional analysis, Sobolev spaces and partial differential equations. Universitext. Springer, New York, 2011. xiv+599 pp. I think this theorem was first proved by Brezis, @DavidMitra – Tomás Dec 23 '12 at 0:32
@Tomas, Can you go further please. – Deepak Dec 23 '12 at 0:32
@Deepak, I finished the proof and if you want to know how to extract the subsequence i have made a citation in the comments. – Tomás Dec 23 '12 at 0:39
Thanks @Tomás.. – Deepak Dec 23 '12 at 0:50

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