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For simplicity, consider the bounded linear operator $L:\ell^\infty\rightarrow \ell^\infty$ defined by $$(Lx)_k=x_{k+1}-x_k,$$ where here, $\ell^\infty$ consists of all bounded bi-infinite sequences of real numbers (ie. indexed by $\mathbb{Z}$). In this article (author's link from their website here, if you don't have access to the journal; see equation (2.9)), it is implied that the dual operator, $L':\ell^1\rightarrow\ell^1$, is given by $$(L'y)_k=y_k-y_{k+1}.$$ Now, if I compute the dual of $L$, I get $$(L'y)_k=(L'y)e_k=y(Le_k)=y(e_{k-1}-e_k)=y_{k-1}-y_k,$$ which is not what is claimed above. Where is my mistake? Note that $(e_i)_j=\delta_{i,j}$; that is, $e_i$ is sequence with a one in position $i$, and zeroes everywhere else.

A quick reminder: if $L:X\rightarrow Y$ is a bounded linear operator and $X$ and $Y$ are Banach spaces, the dual of $L$, denoted $L':Y'\rightarrow X'$, is the unique bounded linear operator satisfying $$(L'y)x=y(Lx)$$ for all $y\in Y'$ and $x\in X$, where $X'$ and $Y'$ are the continuous duals of $X$ and $Y$.

Edit: It has been pointed out that $\ell^1$ is not the dual of $\ell^\infty$. The question now changes: how does one prove that $L':(\ell^\infty)'\rightarrow(\ell^\infty)'$ is given by the same formal definition as it was above (that is, with $(L'y)_k=y_k-y_{k+1}$)?

Kev C
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  • Is the dual of $\ell^\infty(\mathbb{Z})$ in fact $\ell^1(\mathbb{Z})$? And could you define what you mean by dual operator? – Aweygan Aug 04 '16 at 02:17
  • we have $L e_k = e_{k-1}-e_k$ so $(L^* y)k = y^(L e_k) = y^(e{k-1}-e_k) = y_{k-1}-y_k$. Therefore $L^* e_k = e_{k+1} - e_k$ (and to me the word is adjoint) – reuns Aug 04 '16 at 04:24

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One fatal mistake that that $\ell^1$ is not the dual of $\ell^\infty$. This can be seen because $c_0$ (the space of sequences convergent to $0$) is not reflexive, and thus $\ell^1=(c_0)^*$ is not reflexive, hence $\ell^1\neq(\ell^1)^{**}=(\ell^\infty)^*$.

Aweygan
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  • Thank you for pointing this out... not sure how I missed that... In this case, how would one go about proving the correct result? – Kev C Aug 04 '16 at 02:57
  • @KevinChurch Can you tell me where in this paper they imply this alleged implication? – Aweygan Aug 04 '16 at 03:00
  • On page 7, equation 2.9, an operator $L'$ is formally defined that coincides with the operator I mentioned above (with $A_k\equiv I$, in their paper). They then invoke a few results that imply that $L'$ as defined is the dual of $L$ (defined on page 6, equation 2.6). For example, they cite a result saying that if $L$ is Fredholm, so is $L'$. – Kev C Aug 04 '16 at 03:05