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Let $\{e_{n}(x)\}_{n=0}^{\infty}$ be orthnormal basis of Hilbert space $L^2(\mathbb{R})$. If $\{e_{n}(x)\}_{n=0}^{\infty} \subset L^p(\mathbb{R})$ for some $p\geq 1$, is the $\{e_{n}(x)\}_{n=0}^{\infty}$ Schauder basis for $L^p(\mathbb{R})$?

Any reference?

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One can derive the answer "Not in general" working out suitable variations of the following lemma.

Lemma. Let $X$ be a separable Banach space which is not isomorphic to a Hilbert space, but is continuously and injectively embedded into a Hilbert space $H$. Then $X$ contains a sequence $\{x_i\}_{i=1}^\infty$, which is linearly independent, has dense span in $X$, is an orthonormal sequence in $H$, but is not a Schauder basis in $X$ because the linear spans of finite subsets $\{x_i\}_{i=1}^n$ have indefinitely increasing projection constants in $X$.

Proof: By the Lindenstrauss-Tzafriri theorem [Israel J. Math. 9 (1971), 263–269] we can find a sequence $\{y_i\}_{i=1}^\infty$ in $X$ with the dense linear span satisfying the last condition. Applying the Gram-Schmidt ortonormalization, we get the desired sequence $\{x_i\}_{i=1}^\infty$.

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  • $\begingroup$ Mikhail, good answer for a student who asks a question that the student probably should have worked on before asking. More challenging is to have no reordering be a Schauder basis. $\endgroup$
    – Bill Johnson
    Commented Jan 3, 2018 at 23:32
  • $\begingroup$ Hm, is it just orthonormal sequence or an orthonormal basis in $H$? $\endgroup$
    – Fedor Petrov
    Commented Jan 3, 2018 at 23:50
  • $\begingroup$ @Fedor Petrov In this setting it will be an orthonormal sequence only, since I did not require that $X$ is dense in $H$. To get an answer to the Question one has to work because the "continuously embedded" assumption is not satisfied for $L^2(\mathbb{R})$ and $L^p(\mathbb{R})$, but it is satisfied for certain subspaces. I just wanted to show one of the ways to achieve non-Schauderness. $\endgroup$
    – Mikhail Ostrovskii
    Commented Jan 4, 2018 at 0:41
  • $\begingroup$ @BillJohnson It seems that it's not true in general. But does there exist an orthonormal basis in $L^2(\mathbb{R})$ that is also a Schauder basis in $L^p(\mathbb{R})$? How about the case when $ L^2(\mathbb{R})$ and $L^p(\mathbb{R})$ are replaced by $L^2(\mathcal{O})$ and $L^p(\mathcal{O})$ , where $\mathcal{O}$ is a domain with smooth boundary? $\endgroup$
    – ze min jiang
    Commented Sep 22, 2023 at 14:31
  • $\begingroup$ Sure. The Haar system is a basis in $L^p(0,1)$ for all $1\le p < \infty$. Piece Haar systems on disjoint intervals to get a basis on $L^p(-\infty,\infty)$. $\endgroup$
    – Bill Johnson
    Commented Sep 22, 2023 at 19:20
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I am afraid that trigonometric system (multiplied by characteristic functions of the segments $[2\pi k,2\pi(k+1)]$ to become a basis in $L^2$) is not a basis in $L^1$.

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  • $\begingroup$ And for some p>1? $\endgroup$
    – Nebojša Đurić
    Commented Jan 3, 2018 at 21:38
  • $\begingroup$ Trigonometric system is a Schauder basis for all $L^p,p>1$. Probably we may check other specific example like Walsh. $\endgroup$
    – Fedor Petrov
    Commented Jan 3, 2018 at 23:19
  • $\begingroup$ A uniformly bounded orthonormal system that is unconditional in $L^p$ must be equivalent to the unit vector basis of $\ell^2$ (e.g., use type-ecotype theory), so a basis of characters cannot be unconditional. Therefore there is a permutation of any character basis for $L^p$ that is not a basis for $L^p$ when $p\not=2$. $\endgroup$
    – Bill Johnson
    Commented Jan 4, 2018 at 13:08

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