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I wonder if it is possible to have a dense subspace $U \subseteq L^2$ that is disjoint to $L^p$ for some $p\neq 2$. I would expect that such $U$ exists, but I'm stuck finding an example.

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  • $\begingroup$ Other than 0... $\endgroup$ May 2, 2022 at 11:20
  • $\begingroup$ If you mean disjoint except for $\{ 0 \}$ then yes, that's my definition of disjoint spaces. $\endgroup$ May 2, 2022 at 11:23
  • $\begingroup$ Take Sobolev spaces that embed into $L^2$ but not into $L^p$ in case $p>2$ $\endgroup$
    – daw
    May 2, 2022 at 11:56
  • $\begingroup$ Isn't he asking for $L^p \cap U = \{0\}$ but $\bar{U} = L^2$? $\endgroup$
    – Bob
    May 2, 2022 at 11:57
  • $\begingroup$ @Bob is right, this is what I'm asking for. In any case, I don't see in which way Sobolev spaces give an example to any related question, since then all involved spaces are never pairwise disjoint. $\endgroup$ May 2, 2022 at 12:01

2 Answers 2

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I have tried to solve the problem for the interval $(0,1).$ For $-{1\over 2}<t<-{1\over 3}$ consider the functions $f_t(x)=x^{t}.$ Then $f_t\in L^2(0,1)\setminus L^3(0,1).$ Let $$U={\rm span}\, \left \{f_t\,:\, -{1\over 2}<t<-{1\over 3}\right\}$$ Any function $f\in U$ does not belong to $L^3(0,1).$ Indeed, consider a linear combination $$h:=\lambda_1f_{t_1}+\lambda_2f_{t_2}+\ldots +\lambda_nf_{t_n},\qquad -{1\over 2}<t_1<t_2<\ldots <t_n<-{1\over 3}, \ \lambda_j\neq 0$$ Then $$h(x)\approx \lambda_1f_{t_1}(x),\qquad x\to 0^+$$ Hence $h\notin L^3(0,1).$

Assume for a contradiction that $U$ is not dense in $L^2(0,1).$ There exists $g\in L^2(0,1)$ such that $$ \int\limits_0^1x^{t}g(x)\,dx =0,\qquad -{1\over 2}<t<-{1\over 3}$$ Consider the function $$F(z)=\int\limits_0^1e^{z\log x}g(x)\,dx,\qquad \Re z> -{1\over 2}$$ The function $F(z)$ is holomorphic as the uniform limit (for $\Re z\ge -{1\over 2}+\varepsilon$) of holomorphic functions $$F_\delta(z)=\int\limits_{\delta}^1e^{z\log x}g(x)\,dx,\qquad \delta\to 0^+$$ Indeed, $$\displaylines{|F(z)-F_\delta(z)|\le \int\limits_0^\delta e^{\Re z\, \log x}|g(x)|\,dx\\ = \int\limits_0^\delta x^{\Re z}\,|g(x)|\,dx \le \left (\int\limits_0^\delta x^{2\Re z}\,dx\right )^{1/2}\,\|g\|_2}$$ The function $F(z)$ vanishes on the interval $(-1/2,-1/3).$ Therefore it vanishes for any $z,$ $\Re z> -1/2.$ In particular $$F(n)=\int\limits_0^1t^ng(t)\,dt =0,\quad n\in \mathbb{N}_0$$ By the Weierstrass theorem we get $g=0.$

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  • $\begingroup$ Really nice and completely self-contained example, thanks a lot! $\endgroup$ May 3, 2022 at 9:30
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This is the combined result of a previous incompletely justified answer of mine and Giuseppe Negro's suggestion to use Wiener's Tauberian Theorem (https://en.wikipedia.org/wiki/Wiener%27s_Tauberian_theorem).

The example takes place in $L^2(\mathbb R)$.

Let $p$ be any real number in the interval $(-1/2,-1/3]$, and for every $a$ in $\mathbb R$, let $$ f_a(x)=\left\{\matrix{ (x-a)^p, & \text {if } x\in [a,a+1], \cr 0, & \text {otherwise}.\hfill }\right. $$ Note that the $f_a$ are all translates of $f_0$. Observe also that the choice of $p$ ensures that each $f_a$ lies in $L^2(\mathbb R)$ but not in $L^3(\mathbb R)$. Moreover, any function $g$ in $L^2(\mathbb R)$ coinciding with $f_a$ on some interval $(a, a+\varepsilon )$ will also fail to be in $L^3(\mathbb R)$. This said, consider the space $$ U=\text{span}\{f_a: a\in \mathbb R\}. $$ Any nonzero linear combination $$ g=\sum_{i=1}^n \lambda _i f_{a_i}, $$ where we may assume the $\lambda _i$ are nonzero and the $a_i$ are increasing, will coincide with $\lambda _1 f_{a_1}$ on the interval $[a_1,a_2)$, so $g$ will not be in $L^3(\mathbb R)$. This shows that $U\cap L^3(\mathbb R) = \{0\}.$

Next we use Wiener's Tauberian theorem to prove that $U$ is dense. For this all we need to show is that the Fourier fransform of $f_0$ does not vanish on a set of positive measure. But this is clear since $f_0$ is compactly supported and hence $\hat{f_0}$ is analytic.

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    $\begingroup$ This reminds me of the Tauberian theorem of Wiener. $\endgroup$ May 2, 2022 at 16:36
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    $\begingroup$ Well, @GiuseppeNegro since $f_0$ is compactly supported, I suppose it's Fourier transform is analytic, hence cannot vanish on a set on positive measure. Am I right? $\endgroup$
    – Ruy
    May 2, 2022 at 18:02
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    $\begingroup$ Yes you are (I upvoted your previous comment to acknowledge this). Why don't you edit your answer to make it into a complete one? $\endgroup$ May 2, 2022 at 18:34
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    $\begingroup$ @GiuseppeNegro, done! Thanks. $\endgroup$
    – Ruy
    May 2, 2022 at 18:52
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    $\begingroup$ @Ruy Would you mind replacing $\alpha$ by another symbol, as while reading your beatiful solution on the mobile phone it is hard to distinguish $a$ from $\alpha.$ $\endgroup$ May 2, 2022 at 19:27

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