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Every Banach space $X$ is canonically, isometrically embedded in its bidual $X^{**}$. But it is not always $1$-complemented in the bidual: for example, there is no projection from $\ell_\infty$ onto $c_0$, although $c_0^{**}=\ell_\infty$.

Let $\mathbb{H}$ be a separable Hilbert space, then the bounded operators $\mathbb{B}(\mathbb{H})$ form a Banach space. How to show that this space is $1$-complemented in the bidual, meaning there exists a linear contraction $\tau:\mathbb{B(H)^{**}}\rightarrow \mathbb{B(H)}$ such that $\tau(T)=T$ for $T$ in $\mathbb{B(H)}$?

David Lee
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For every normed space $X$, the dual $X^*$ is $1$-complemented in $X^{***}$. Indeed, let $i:X\to X^{**}$ be the canonical embedding; then its adjoint $i^*$ is a projection of norm $1$ of $X^{***}$ to $X^*$. Simply put, it takes a functional $\phi:X^{**}\to \mathbb{C}$ and composes it with $i$.

In particular, the above applies to $\mathbb{B}(\mathbb{H})$, which is the dual of the space of trace-class operators.