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Let $X$ a set, $M$ a $\sigma$–algebra and $\mu$ a measure. We have known that $L^p(X,M,\mu)$ is a Banach space if $p\geq 1$. In particular when $p=2$, $L^p(X,M,\mu)$ is a Hilbert space,(which are classical theorems.)

My question is, if $L^p(X,M,\mu)$ is a Hilbert space can we conclude that $p=2$?

Or maybe we will add some conditions on $X$ or $(X,M)$ or $(X,M,μ)$?

Help me, please. This question has bothered me for a long time.

hardmath
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David Chan
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  • I don't think this should have been closed, given the OPs comments to my answer, it is not clear that they see how to make the jump that the Parallelogram law doesn't hold in $L^p$ for $p\neq 2$, and thus the linked "duplicate" is insufficient. – jxnh Mar 16 '15 at 14:30

2 Answers2

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A Banach space is Hilbert if and only if it satisfies the parallelogram identity. Does this answer your question?

Squirtle
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  • I'm sorry for that. As a matter of fact, I want to clear the results about this question"L^p is a Hilbert space concludes p=2" – David Chan Mar 16 '15 at 03:03
  • I found no question called, "L^p is a Hilbert space concludes p=2" – Squirtle Mar 16 '15 at 03:04
  • The question is "if L^p(p \geq 1)(has been a Banach space) is a Hilbert space,then p must be equal to 2". – David Chan Mar 16 '15 at 03:08
  • Did you look at the "Duplicate question" I linked to above (under your posted question)? – Squirtle Mar 16 '15 at 03:10
  • Yes, I have seen that, but my question is not compatible with that question. My question is a specific question.(may be wrong),in that question,and the answers explain if a "abstract" Banach space is a Hilbert space, then this Banach space must satisfy which condition. I need the way to realize this condition or how to apply it specifically. – David Chan Mar 16 '15 at 03:21
  • The proof to "Hilbert if and only if Parallelogram Law holds" is tough. However, assuming that result, it is easy to show that $L^p$ being Hilbert implies $p=2$. Because:

    http://math.stackexchange.com/questions/216306/ell-p-is-hilbert-space-if-and-only-if-p-2

     – Squirtle Mar 16 '15 at 03:27
  • I didn't give this answer before because you really should have tried to use the fact I put in my first answer and attempt the algebra for yourself (because its rather straight forward.... while the equivalence of Hilbert and Parallelogram is not) – Squirtle Mar 16 '15 at 03:29
  • :thanks, but you really do not know my idea. L^p(X) is not ℓ^p, ℓ^p is just a special situation when X=the set of natural numbers. – David Chan Mar 16 '15 at 03:35
  • I believe the argument can easily be made generally; but I'm going to sleep. It probably involves using a characteristic function of some sort. – Squirtle Mar 16 '15 at 03:51
  • Yes,please help me if you have time.The question has really bothered me for a long time. – David Chan Mar 16 '15 at 03:57
  • Have a good night. – David Chan Mar 16 '15 at 03:57
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Suppose your measure space has two disjoint measurable subsets $A,B$ both of finite measure. Then there is a map $f:\mathbb{R}^2 \to L^p(X)$ given by sending $(a,b)$ to the function $$[f(a,b)](x) = \begin{cases} \frac{a}{\mu(A)^\frac{1}{p}} \text{ if } x \in A, \\ \frac{b}{\mu(B)^\frac{1}{p}} \text{ if } x \in B \\ 0 \text{ otherwise.} \end{cases} $$

You can check that $\| f(a,b) \| = (a^p + b^p)^{1/p}$. On the other hand, if $L^p(X,\mu)$ is a Hilbert space then the parallelogram law holds, which would imply the parallelogram law for 2-d space with $p$-norm. Thus one is reduced to showing that $\mathbb{R}^2$ is a Hilbert space only under the $p = 2$ norm.

Edit: I should point out that the restriction I give at the beginning fails only (as far as I can see) in cases where all the $L^p$ (save maybe the uninteresting $L^\infty$ case) are one or zero dimensional.

Edit2: If a space has no sets of non-zero finite measure, then every $L^p$ function will be almost everywhere zero except for $p = \infty$, so yeah in this case we have $L^p$ a Hilbert space for some $p \neq 0$ , but only because it is zero.

Now suppose we have sets of finite non-zero measure but no disjoint pair. Suppose $A$ has non-zero finite measure, then this means that for any $B$ with non-zero finite measure we would have: $\mu(B) = \mu(A \cap B) = \mu(A)$. It follows from this (though I don't want to write out al the details) that $L^p(X)$ will be one dimensional, with the elements just rescalings of the indicator of the set $A$.

jxnh
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  • In fact, I'm familiar with this incomplete proof."Suppose your measure space has two disjoint measurable subsets A,B both of finite measure."How to realize it for any space? – David Chan Mar 16 '15 at 03:05
  • Oh no,I find you incomplete proof is a little different from me.Er, why"which would imply the parallelogram law for 2-d space with p -norm."And another question, f is a linear map? – David Chan Mar 16 '15 at 04:16
  • @DavidChan Yes, f is pretty clearly linear. To be honest, you should be more than comfortable with eyeballing what formulas give linear maps before tackling Hilbert/Banach spaces. (as I read this I realize I overloaded the letter f, so after the edits maybe its better).

    Two, given an injective linear map $f: E \to F$ with $E$ a vector space and $F$ a normed vector space, there is a norm on $E$ given by $|v|_E = | f(v) |_F$. That's what I'm doing here. And if the norm in $F$ satisfies a parallelogram law you can check that the norm we just defined on $E$ does as well.

     – jxnh Mar 16 '15 at 06:33
  • Well, you are so nice. But this question "Suppose your measure space has two disjoint measurable subsets A,B both of finite measure"? Did every measure space has two disjoint finite measurable subsets? – David Chan Mar 16 '15 at 06:51
  • @DavidChan I've added some more comments about the restriction. – jxnh Mar 16 '15 at 07:00
  • It follows from this (though I don't want to write out al the details) that L^p (X) will be one dimensional;I really can't find how to conclude that L p (X) is one dimensional. – David Chan Mar 16 '15 at 09:09