## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorTobyBartels
• CommentTimeAug 17th 2011

I added some more to Lebesgue space about the cases where $1 \lt p \lt \infty$ fails.

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeAug 17th 2011
• (edited Aug 17th 2011)

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.

• CommentRowNumber3.
• CommentAuthorTobyBartels
• CommentTimeAug 18th 2011

Yeah, I remembered that wrong. Fixed.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeAug 18th 2011
• (edited Aug 18th 2011)

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.

• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeAug 18th 2011

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 ‘${|x|} = {|y|}$’.
• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeAug 18th 2011

Thanks for the tip! “Who knew?”

• CommentRowNumber7.
• CommentAuthorTobyBartels
• CommentTimeAug 18th 2011

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

• CommentRowNumber8.
• CommentAuthorzskoda
• CommentTimeAug 18th 2011

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

• CommentRowNumber9.
• CommentAuthorTobyBartels
• CommentTimeAug 18th 2011
• (edited Aug 18th 2011)

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

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeAug 18th 2011
• (edited Aug 18th 2011)

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.

• CommentRowNumber11.
• CommentAuthorTobyBartels
• CommentTimeAug 18th 2011

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

• CommentRowNumber12.
• CommentAuthorDmitri Pavlov
• CommentTimeMay 27th 2022

## 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 $i$.