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Let $j_X:X\rightarrow X^{**}$ denote the canonical embedding.

I've read several articles where it is assumed that the reader is familiar with the idea that there is a norm one projection from $X^{***}$ to $X^*$. More precisely, $P:=j_{X^*}(j_X)^*$ seems to be that projection. However, I couldn't find any material that would explain it in detail, and I can't really figure out how this operator works on my own.

If we take $x^{***} \in X^{***}$, then what exactly happens to it when we apply $P$, in other words, what kind of an operator is $j_{X^*}(j_X)^*x^{***}: X^{**} \rightarrow \mathbb{K}$?

Ksenia
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2 Answers2

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$P(x^{***})(x) = x^{***}(j_X(x))$

Emanuele Paolini
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This is an exercise in writing down the definitions. Recall the general definitions of $j_X$, and of the dual operator. Apply this here. In the end, you get

$$(j_{X^*} \circ (j_X)^*)(v)(u) = u(v \circ j_X)$$

for $v \in X^{***}$ and $u \in X^{**}$.

Martin Brandenburg
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