$L^p$ is reflexive for $1<p<2$, Brezis proof

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Theorem 4.10 [pages 95-97] of Brezis proves that $L^p$ is reflexive for $p\in(1,\infty)$.
The last statement of the proof says that since $T(L^p)$ is reflexive, with $T$ being an isometry, so is $L^p$. Why is it true?

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In general, if $E, \ F$ are isomorphic normed vector spaces, then $F$ is reflexive $\Leftrightarrow$ $E$ is reflexive. Since $T$ is an isomorphism (onto $T(L^p)$), it follows that $L^p$ is reflexive.