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In this question the norm of $L^{P}[0,1]$ is denoted by $\parallel . \parallel _{p}$.

Let $p$ and $q$ be two arbitrary real numbers with $2<p<q$.

Assume that $S$ is a subvector space of $L^{q}[0, 1]$ such that the identity operator $\text{Id.}: (S, \parallel . \parallel _{p}) \to (S, \parallel . \parallel _{q})$ is a bounded operator. Does this implies that $S$ is a finite dimensional space?

If I am note mistaken, this is proved for $p=2, q=\infty$, by Grothendieck.

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No. The condition implies that the subspace is isomorphic to a Hilbert space. In fact, Kadec and Pelczynski proved that a subspace of $L_p$, $2<p<\infty$, is closed in $L_r$ for some $r<p$ if and only if the subspace is isomorphic to a Hilbert space.

Kadec, M. I.; Pełczyński, A. Bases, lacunary sequences and complemented subspaces in the spaces Lp. Studia Math. 21 1961/1962 161–176.

Or look at the book by Albiac and Kalton.

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  • $\begingroup$ Prof. Johnson, Thank you very much for your answer. So we need to know that for every $p>2$, $L_{p}$ contains an infinite dimensional Hilbert space, isometricaly. is it obvious?(I apologize if this question is elementary) $\endgroup$
    – Ali Taghavi
    Commented Jan 22, 2014 at 7:50
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    $\begingroup$ @AliTaghavi: What you need to know is that $L_p$ contains an infinite dimensional Hilbert space, isomorphically. It is not obvious, but it is true and classical. The usual proof is that Khintchine's inequality shows that the span of the Rademacher functions is such a space. $\endgroup$
    – Mark Meckes
    Commented Jan 22, 2014 at 12:02
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    $\begingroup$ Or, if you want an isometric infinite dimensional Hilbert space in $L_p$, consider the closed span of an independent sequence of $N(0,1)$ random variables. $\endgroup$
    – Bill Johnson
    Commented Jan 22, 2014 at 14:23
  • $\begingroup$ Yes, of course, silly of me to forget that. I guess I got too hung up on the fact that an isometric embedding wasn't strictly needed here. Incidentally, it's worth noting that even though an isometric embedding isn't necessary, it's better than the isomorphic embedding I suggested, for which the norm of the embedding blows up when $p \to \infty$. $\endgroup$
    – Mark Meckes
    Commented Jan 23, 2014 at 8:29

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