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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](http://math.stackexchange.com/questions/18430/is-this-an-inner-product-on-l1) post.2012-12-23

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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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    It looks ok to me; but I think using Riesz is a bit of a "sledgehammer" (See the link in my comment to the OP)...2012-12-23