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1. • CommentRowNumber1.
   • CommentAuthorTobyBartels
   • CommentTimeAug 17th 2011
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexI added some more to [[Lebesgue space]] about the cases where $1 \lt p \lt \infty$ fails.

   I added some more to Lebesgue space about the cases where $!A = \sum_{E: V} (E \to A}1 \lt p \lt \infty$ fails.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 17th 2011
   • (edited Aug 17th 2011)
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   Author: Todd_Trimble
   Format: MarkdownItexHi Toby. Could you check again the asserted local convexity in the cases $0 \lt p \lt 1$? Because the Hahn-Banach theorem implies that the dual of a locally convex TVS is non-zero, whereas it is known that $L^p$ for this range of $p$ typically has zero dual. See also <a href="http://en.wikipedia.org/wiki/Lp_space#Lp_for_0_.3C_p_.3C_1">this section</a> from Wikipedia.

   Hi Toby. Could you check again the asserted local convexity in the cases $0 \lt p \lt 1$? Because the Hahn-Banach theorem implies that the dual of a locally convex TVS is non-zero, whereas it is known that $L^p$ for this range of $p$ typically has zero dual. See also this section from Wikipedia.

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeAug 18th 2011
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   Author: TobyBartels
   Format: MarkdownItexYeah, I remembered that wrong. Fixed.

   Yeah, I remembered that wrong. Fixed.

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 18th 2011
   • (edited Aug 18th 2011)
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   Author: Todd_Trimble
   Format: MarkdownItexI added a section on Minkowski's inequality for the case $1 \leq p \leq \infty$, with a proof of my own devising. I don't think it's actually original with me, but I've not seen it in the books I've looked at. The textbook proofs I have seen involve H&ouml;lder's inequality, but without the courtesy of saying what is going on in that proof *conceptually*. I have a page on these issues on my lab, <a href="http://ncatlab.org/toddtrimble/published/p-norms">here</a>.

   I added a section on Minkowski’s inequality for the case $1 \leq p \leq \infty$, with a proof of my own devising. I don’t think it’s actually original with me, but I’ve not seen it in the books I’ve looked at. The textbook proofs I have seen involve Hölder’s inequality, but without the courtesy of saying what is going on in that proof conceptually. I have a page on these issues on my lab, here.

5. • CommentRowNumber5.
   • CommentAuthorTobyBartels
   • CommentTimeAug 18th 2011
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   Author: TobyBartels
   Format: MarkdownItexCool! By the way, when putting norms and absolute values in itex, it looks a lot better if you put each one inside braces. Compare: * `|x| = |y|` produces ‘$|x| = |y|$’; * `{|x|} = {|y|}` produces ‘${|x|} = {|y|}$’.

   Cool!

   By the way, when putting norms and absolute values in itex, it looks a lot better if you put each one inside braces. Compare:

   • |x| = |y| produces ‘$|x| = |y|$’;
   • {|x|} = {|y|} produces ‘${|x|} = {|y|}$’.
6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 18th 2011
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   Author: Todd_Trimble
   Format: MarkdownItexThanks for the tip! "Who knew?"

   Thanks for the tip! “Who knew?”

7. • CommentRowNumber7.
   • CommentAuthorTobyBartels
   • CommentTimeAug 18th 2011
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   Author: TobyBartels
   Format: MarkdownItexNow that I'm looking at this on my phone, they look identical (and, unusually, better). That's weird!

   Now that I’m looking at this on my phone, they look identical (and, unusually, better). That’s weird!

8. • CommentRowNumber8.
   • CommentAuthorzskoda
   • CommentTimeAug 18th 2011
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   Author: zskoda
   Format: MarkdownItexMy memory is that between $0$ and $1$ we have still Frechet spaces.

   My memory is that between $0$ and $1$ we have still Frechet spaces.

9. • CommentRowNumber9.
   • CommentAuthorTobyBartels
   • CommentTimeAug 18th 2011
   • (edited Aug 18th 2011)
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexBut Fréchet spaces are locally convex, giving Todd's objection again. (I seem to recall that $F$-spaces have sometimes been called "Fréchet spaces".)

   But Fréchet spaces are locally convex, giving Todd’s objection again. (I seem to recall that $F$-spaces have sometimes been called “Fréchet spaces”.)

10. • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 18th 2011
    • (edited Aug 18th 2011)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexToby #9 is right. There is some terrible terminological confusion here. I grew up with the meaning of Fr&eacute;chet space as including local convexity (which rules out $L^{1/2}$), but apparently some people use it to mean what is called <a href="http://en.wikipedia.org/wiki/F-space">F-space</a>. A Fr&eacute;chet space in my meaning is a TVS whose topology is given by a countable family of seminorms under which the TVS becomes a complete metric space (including the axiom that $d(x, y) = 0$ implies $x = y$). If you have a Fr&eacute;chet space given by a single seminorm, then that is a norm and you get a Banach space.

    Toby #9 is right. There is some terrible terminological confusion here. I grew up with the meaning of Fréchet space as including local convexity (which rules out $L^{1/2}$), but apparently some people use it to mean what is called F-space.

    A Fréchet space in my meaning is a TVS whose topology is given by a countable family of seminorms under which the TVS becomes a complete metric space (including the axiom that $d(x, y) = 0$ implies $x = y$). If you have a Fréchet space given by a single seminorm, then that is a norm and you get a Banach space.

11. • CommentRowNumber11.
    • CommentAuthorTobyBartels
    • CommentTimeAug 18th 2011
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexI clarified that $L^0$ is not even an $F$-space.

    I clarified that $L^0$ is not even an $F$-space.

12. • CommentRowNumber12.
    • CommentAuthorDmitri Pavlov
    • CommentTimeMay 27th 2022
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    Author: Dmitri Pavlov
    Format: MarkdownItexAdded: ## Notation For historical reasons (starting with the original paper by Riesz), the exponent $p$ is traditionally taken to be the [[reciprocal]] of the “correct” exponent. If we take $M^p=L^{1/p}$, the spaces $M^p$ form a $\mathbf{C}$-[[graded algebra]], where $\mathbf{C}$ denotes [[complex numbers]]. This is a conceptual explanation for the appearance of formulas like $1/p+1/q=1/r$ in [[Hölder's inequality]]. In [[differential geometry]], the notion of [[density]] does use the “correct” grading. In the [[Tomita–Takesaki theory]], the parameter $t$ for [[modular automorphism group]] is almost the “correct” grading, except that it is multiplied by the [[imaginary unit]] $i$. <a href="https://ncatlab.org/nlab/revision/diff/Lebesgue+space/23">diff</a>, <a href="https://ncatlab.org/nlab/revision/Lebesgue+space/23">v23</a>, <a href="https://ncatlab.org/nlab/show/Lebesgue+space">current</a>

    Added:

    Notation

    For historical reasons (starting with the original paper by Riesz), the exponent $p$ is traditionally taken to be the reciprocal of the “correct” exponent.

    If we take $M^p=L^{1/p}$, the spaces $M^p$ form a $\mathbf{C}$-graded algebra, where $\mathbf{C}$ denotes complex numbers.

    This is a conceptual explanation for the appearance of formulas like $1/p+1/q=1/r$ in Hölder’s inequality.

    In differential geometry, the notion of density does use the “correct” grading.

    In the Tomita–Takesaki theory, the parameter $t$ for modular automorphism group is almost the “correct” grading, except that it is multiplied by the imaginary unit $i$.

    diff, v23, current