# dense subspace of $L^2$ that is disjoint to $L^p$

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.

• Other than 0... May 2, 2022 at 11:20
• If you mean disjoint except for $\{ 0 \}$ then yes, that's my definition of disjoint spaces. May 2, 2022 at 11:23
• Take Sobolev spaces that embed into $L^2$ but not into $L^p$ in case $p>2$
– daw
May 2, 2022 at 11:56
• Isn't he asking for $L^p \cap U = \{0\}$ but $\bar{U} = L^2$?
– Bob
May 2, 2022 at 11:57
• @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. May 2, 2022 at 12:01

I have tried to solve the problem for the interval $$(0,1).$$ For $$-{1\over 2} consider the functions $$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} Any function $$f\in U$$ does not belong to $$L^3(0,1).$$ Indeed, consider a linear combination $$h:=\lambda_1f_{t_1}+\lambda_2f_{t_2}+\ldots +\lambda_nf_{t_n},\qquad -{1\over 2} Then $$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 $$\int\limits_0^1x^{t}g(x)\,dx =0,\qquad -{1\over 2} Consider the function $$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.$$

• Really nice and completely self-contained example, thanks a lot! May 3, 2022 at 9:30

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.

• This reminds me of the Tauberian theorem of Wiener. May 2, 2022 at 16:36
• 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?
– Ruy
May 2, 2022 at 18:02
• 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? May 2, 2022 at 18:34
• @GiuseppeNegro, done! Thanks.
– Ruy
May 2, 2022 at 18:52
• @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 $\alpha.$ May 2, 2022 at 19:27