Schauder basis $L^p(\mathbb{R})$ 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?
 A: 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 $\{x_i\}_{i=1}^\infty$.
A: 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 $L^1$.
