meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 3 months |
| seen | Jun 18 '12 at 21:16 | |
| stats | profile views | 78 |
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Mar 13 |
awarded | Popular Question |
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Jan 24 |
awarded | Yearling |
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Jun 8 |
awarded | Caucus |
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May 11 |
accepted | Question on singular measures and absolute continuity |
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May 11 |
revised |
A Question on Convergence In $L^p$ added 1 characters in body |
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May 11 |
revised |
A Question on Convergence In $L^p$ added 1 characters in body |
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May 11 |
revised |
A Question on Convergence In $L^p$ added 1 characters in body |
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May 9 |
comment |
An application of Riesz Representation Theorem @leo: then $f$ would have to 1. right? |
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May 9 |
comment |
An application of Riesz Representation Theorem @leo: I don't get your point. Do you mean $\int|g-h|=0$ implies $g=h$? |
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May 8 |
comment |
How to show a sequence is not uniformly integrable @DavidMitra: How does one find such an $n$? |
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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. |
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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 $fg\notin L^1$. |
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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 |
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May 7 |
comment |
An application of Riesz Representation Theorem @t.b. Because $\int f(g-h) =0$ |
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May 7 |
comment |
An application of Riesz Representation Theorem @t.b. Am I right in saying that $f(g-h)=0$ so that $g=h$? |
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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. |
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May 6 |
asked | An application of Riesz Representation Theorem |
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May 5 |
comment |
How to show that $f=0$ a.e. on $[0,1]\times [0,1]$. Thanks for your answer. Please can you expand more on your answer. I'm not familiar with Lebesgue differentiation theorem. |
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May 5 |
comment |
How to show that $f=0$ a.e. on $[0,1]\times [0,1]$. @BenDerrett: I don't understand what you mean. |
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May 5 |
asked | How to show that $f=0$ a.e. on $[0,1]\times [0,1]$. |
