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The map $T:\ell^1\to (\ell^\infty)', (Tx)(y)=\sum_{n=1}^\infty x_ny_n$ is isometric, but not surjective.

According to my book it is easy to prove that $T$ is isometric, but I don't quite know how to show this. I think we have to show that $$\left|T(x)(y)\right|=\|y\|_{\ell^\infty}$$ but I don't really know how to even start this.

dinosaur
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    No, that $T$ is isometric means $\lVert Tx\rVert_{(\ell^\infty)^\ast} = \lVert x\rVert_{\ell^1}$, and that means $$\sup { \lvert (Tx)(y)\rvert : \lVert y\rVert_{\ell^\infty} \leqslant 1} = \lVert x\rVert_{\ell^1}.$$ – Daniel Fischer Jun 04 '14 at 18:20

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What we have to show is that the norm of the linear functional $T(x)$ is $\lVert x\rVert_1$. This means that $$\lVert T(x)\rVert:=\sup_{y\in \ell^\infty,y\neq 0}\frac{|T(x)(y)|}{\lVert y\rVert_\infty}=\lVert x\rVert_1,$$ but not that equality $\frac{|T(x)(y)|}{\lVert y\rVert_\infty}=\lVert x\rVert_1$ holds for any $y$.

Notice that $|T(x)(y)|\leqslant \sup_j|y_j|\cdot\sum_{k=1}^{+\infty}|x_k|=\lVert x\rVert_1\cdot \lVert y\rVert_\infty$, hence $\lVert T(x)\rVert\leqslant \lVert x\rVert_1$.

For the reverse equality, consider $y$ where $y_k=1$ if $x_k\gt 0$, $y_k=-1$ if $x_k$ is negative and $y_k=0$ if $x_k=0$.

Davide Giraudo
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