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I added some more to Lebesgue space about the cases where $1 \lt p \lt \infty$ fails.
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.
Yeah, I remembered that wrong. Fixed.
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.
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 ‘${|x|} = {|y|}$’.Thanks for the tip! “Who knew?”
Now that I’m looking at this on my phone, they look identical (and, unusually, better). That’s weird!
My memory is that between $0$ and $1$ we have still Frechet 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”.)
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.
I clarified that $L^0$ is not even an $F$-space.
Added:
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 $i$.
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