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seen Jun 18 '12 at 21:16

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awarded  Notable Question
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awarded  Yearling
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May
11
accepted Question on singular measures and absolute continuity
May
11
revised A Question on Convergence In $L^p$
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May
11
revised A Question on Convergence In $L^p$
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May
11
revised A Question on Convergence In $L^p$
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May
9
comment An application of Riesz Representation Theorem
@leo: then $f$ would have to 1. right?
May
9
comment An application of Riesz Representation Theorem
@leo: I don't get your point. Do you mean $\int|g-h|=0$ implies $g=h$?
May
8
comment How to show a sequence is not uniformly integrable
@DavidMitra: How does one find such an $n$?
May
8
comment If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$?
Thanks very very much.
May
7
comment If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$?
sorry if I wasn't clear before. I was wondering if you could expand on the derivation of the contradiction. for example why $g=\sum\limits_{k=1}^\infty2^{-k}g_k\|g\|_{L^q}$ is less than one and why it implies that $fg\notin L^1$.
May
7
comment If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$?
@robjohn: do you mind elaborating on your proof a bit. the part where you showed that $\|fg\|_{L^1} \leq C \|g\|_{L^q}$? Thanks
May
7
comment An application of Riesz Representation Theorem
@t.b. Because $\int f(g-h) =0$
May
7
comment An application of Riesz Representation Theorem
@t.b. Am I right in saying that $f(g-h)=0$ so that $g=h$?
May
7
comment An application of Riesz Representation Theorem
@t.b. Thanks for the link. Could you please help me with the conclusion. I don't quite get how showing that $\| f g\|_1 \leq M\|f\|_2$ shows that $g\in L^2[0,1]$. Thanks.