Let $i :X \rightarrow X^{**}$ and $j : X^* \rightarrow X^{***}$ be natural embeddings. Then $Q := j \circ i^*$ is a projection with $Q(X^{∗∗∗}) = X^∗$

Question

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I would to like to ask two questions in line 3 of the given proof of the proposition.

My first question would be how does $Q\circ j = j$ imply image of $Q = j(X^*)$. The image of $Q$ equals to $Q(X^{***})$ but I couldn't find such a equality which links $X^*$ and $X^{***}$ together.

My second question would be how could I prove that image of $Q = X^*$ if the image of $Q = j(X^*)$.

Any hints or help would be greatly appreciated. Thank you.

Answer 1

For your first question, $Q\circ j=j$ seems useless. Just say$$Q(X^{***})=j\circ i^*(X^{***})=j(X^*),$$the last equality being due to the fact that $i^*$ is onto since $i^*\circ j=Id_{X^*}$ (or: since $i$ is 1-to-1).

For your second question, the author just makes his original sentence more correct when he replaces his (invalid) ${\rm im}(Q)=X^*$ by his (correct) ${\rm im}(Q)=j(X^*)$. His first rough formulation was motivated by the fact that $j$ is an embedding.

And here is my answer to Surb's new question (in the comments below, about $i^*\circ j=Id_{X^*}$):

• The embedding $i:X\to X^{**}$ (where $X^*$ denotes the topological dual space of $X$, with the strong topology) is defined by $(i(x))(\varphi)=\varphi(x)$.
• Its transpose $i^*:X^{***}\to X^*$ is defined by $i^*(\psi)=\psi\circ i$.
• The embedding $j:X^*\to X^{***}$ is defined the same way as $i$: $(j(\varphi))(\rho)=\rho(\varphi)$. Now,$$[(i^*\circ j)(\varphi)](x)=[(j(\varphi))\circ i](x)=(i(x))(\varphi)=\varphi(x)$$(for all $x\in X$ and $\varphi\in X^*$), hence $i^*\circ j=Id_{X^*}$.