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.
 A: 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.
A: 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 $g=0.$
