Let $(g_n)_{n\geq 1}$ be a sequence of i.i.d. complex Gaussian random variables on $[0,1].$ Then it is easy to see that the map $j:\ell_2\to L^p([0,1])$ defined as $je_n=[E(g_n^p)]^{\frac{1}{p}}g_n,n\geq 1$ is an isometry and $j(\ell_2)$ is a closed subset of $L^p([0,1])$ which is also complemented. Let $P:L^p([0,1])\to j(\ell_2)$ be the projection map. Then what can be said about the norm of $P$ ?

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    $\begingroup$ When you claim that the closed span of the Gaussians is complemented in $L^p[0,1]$, doesn't this require $1<p<\infty$? That is, I seem to recall that for $p=1$ we have an uncomplemented subspace. $\endgroup$
    – Yemon Choi
    Commented Aug 17, 2017 at 21:08
  • $\begingroup$ yes! Off course! Sorry I did not mentioned that. $\endgroup$
    – Mathbuff
    Commented Aug 17, 2017 at 21:15
  • $\begingroup$ @ Pietro Majer. I have edited the question. But as I mentioned in the question $j(\ell_2)$ is a closed and complemented subspace of $L^p([0,1]).$ Hence the existence of the bounded projection is clear. Thank you. $\endgroup$
    – Mathbuff
    Commented Aug 18, 2017 at 15:42
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    $\begingroup$ Think about why $\|g_1\|_p$ is an upper bound when $2<p<\infty$, then consult some standard text. I don't have mine where I am now but I think it is in Tomczak-Jaegermann's book or Wojtaszczyk's book; maybe also in Albiac-Kalton. $\endgroup$ Commented Aug 18, 2017 at 19:48
  • $\begingroup$ @Bill. Thank you Bill. I would look into that. Thank you very much. $\endgroup$
    – Mathbuff
    Commented Aug 19, 2017 at 10:26


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