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Let $E$ and $F$ be Banach spaces and $T:E\to F$ a continuous linear operator. Is this following equality correct? $$(T E)''=T''E''$$

Notes:

  • $E''$ is the bidual of $E$
  • $T''$ is the second adjoint of $T$
  • $TE=\{Tu:u\in E\}$
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Yes. To state this more precisely, introduce the canonical inclusions $\rho_E:E\to E''$ and $\rho_F:F\to F''$. The claim follows from the commutative diagram (which I won't draw), $$ T''\circ \rho_E = \rho_F\circ T \tag1$$ (Equal operators have the same range.)

For any $x\in E$ and any $f^*\in F'$, the definition of adjoint operator yields $$ \langle T''\rho_E(x), f^*\rangle = \langle \rho_E(x), T'f^*\rangle = \langle x, T'f^*\rangle = \langle Tx, f^*\rangle = \langle \rho_F(Tx), f^*\rangle $$ where angle brackets mean the natural pairing of a space with its dual. Since $f^*$ was arbitrary, we get $$ T''\rho_E(x) = \rho_F(Tx) $$ which is (1).

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