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Let $\left(l_1\right)^*$ be the dual space to $l_1$. Each $f \in \left(l_1\right)^*$ is a continuous linear functional over $l_1$. There is constant $C \in \Bbb R$ such that $|f(x)|\le C|x|_1, \forall x\in l_1$, (note that we denote the $L^1$-norm of $x\in l_1 $ by $|x|_1$).

Also, we define the norm for $f \in \left(l_1\right)^*$ by$$||f||:= sup \{|f(x)|:\forall x\in l_1 \text {such that }|x|_{1}=1\}$$

For each $\left(b_n\right)\in l_{\infty}$, define its norm by $|(b_n)|_{\infty}:=sup\{a_n:\forall n\}$, where $l_{\infty}$ is the vector space of all bounded real sequences.

Show that there is a continuous vector space isomorphism $M:l_{\infty}\rightarrow \left(l_1\right)^* $ such that $$||M(x)||\le C|x|_{\infty} $$ for some constant $C$.

My thought is to show that as in the finite-dimensional case, check that a bounded sequence $b:=(b_n)$ defines a linear functional $$f_b:(a_1,a_2,...)\in l_1 \mapsto \sum_{n=1}^{\infty} b_na_n \in \Bbb R$$, but I'm not sure how to show this exactly, and stuck on showing the following steps. Could someone give a complete proof of the question please? Any help is appreciated. Thanks a lot.

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You have the right idea. Check that the partial sums $\sum^n_{i=1} |b_na_n|$ are bounded above by $\|b\|_\infty \|a\|_1$ to see that the series you used to define $f$ converges and gives a functional of norm at most $\|b\|_\infty$. To see that $\|f_b\| = \|b\|_\infty$ consider what happens when you apply $f_b$ to basis vectors of $l_1$. That'll tell you that $b \mapsto f_b$ is an isometric linear map of $l_\infty$ into $l_1^*$. To see that it's surjective, fix a functional $f$ in $l_1^*$ and find an element $b$ of $\ell_\infty$ such that $f_b$ and $f$ agree on basis vectors for $l_1$.

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  • $\begingroup$ Do you mean that $\|f_b\| = \|b\|_\infty$ implies that $b \mapsto f_b$ is an isometric linear map of $l_\infty$ into $l_1^*$? $\endgroup$ – nora012 Apr 5 '16 at 19:59
  • $\begingroup$ No, you have to do a simple calculation to check that it's linear (just apply $f_{\lambda b + c}$ and $\lambda f_b + f_c$ to an arbitrary basis vector and compare). Then $\|f_b\| = \|b\|_\infty$ implies that $b \mapsto f_b$ is isometric. $\endgroup$ – Aidan Sims Apr 5 '16 at 21:58
  • $\begingroup$ Did this answer your question? Do you need more detail? $\endgroup$ – Aidan Sims Apr 6 '16 at 22:25

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