# The dual of the direct sum

Let $X$, $Y$, $Z$ normed spaces If X$\cong Y\oplus Z$ why is $X^*\cong Y^*\oplus Z^*$? where $X^*$ is the dual of $X$. For example ${\ell^\infty}^*\cong\ell^1\oplus\mathrm{Null}\;C_0$ so if we take the double dual we find the ${\ell^\infty}^{**}\cong{\ell^1}^*\oplus (\mathrm{Null}\; C_0)^*$. I am not sure I understand why these equalities should hold?

-
Do you understand the corresponding fact for vector spaces? – Qiaochu Yuan Oct 20 '12 at 20:17

Let $X \cong Y \oplus Z$, that is $Y$ and $Z$ can be regarded as closed, complemented subspaces of $X$. Now define $T \colon X^* \to Y^* \oplus Z^*$ by $Tx^* = (x^*|_Y, x^*|_Z)$. Then
• $T$ is bounded, as \begin{align*} \|Tx^*\| &= \|(x^*|_Y, x^*|_Z)\|\\ &\le C \bigl(\|x^*|_Y\| + \|x^*|_Z\|\bigr)\\ &\le 2C\|x^*\|. \end{align*}
• $T$ is one-to-one as for $x^*$ with $Tx^* = 0$ we have for $x \in X$, written as $x = y+z$, that $x^*(x) = x^*|_Y(y) + x^*|_Z(z) = 0+0=0$, so $x^* = 0$.
• $T$ is onto: Let $y^* \in Y$, $z^* \in Z$. Define $x^*\colon X \cong Y \oplus Z \to \mathbb K$ by $x^*(y+z) = y^*(y) + z^*(z)$, then $x^*$ is bounded as the projections onto $Y$ and $Z$ are bounded and $y^*$ and $z^*$ are, so $x^* \in X^*$ and obviously $Tx^*= (y^*,z^*)$.
• $T^{-1}$ is bounded: $T^{-1}(y^*, z^*) = y^*P_Y + z^*P_Z$ ($P_Y$, $P_Z$ denoting the projections) is bounded, as the projections are.
So $T$ is an isomorphism.
We have $X \cong Y \oplus Z$. So $y \mapsto (y,0)$ and $z \mapsto (z,0)$ are continuous maps, with the projections being their inverses. So $Y \oplus \{0\}$ is isomorphic to $Y$ and a closed subspace of $X$. – martini Oct 24 '12 at 20:54