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For $y\in {\ell }^{\infty }$, the formula ${\varphi }_y(x)={\sum }_n{x}_n{y}_n$ defines a bounded functional on ${\ell }^1$ with $\Vert {\varphi }_y\Vert =\Vert y{\Vert }_{\infty }$. Conversely, every $\varphi \in ({\ell }^1{)}^{\ast }$ is represented this way by the bounded sequence $y_n=\varphi (e_n)$. This is an isometric isomorphism of normed spaces.

The continuous dual space of a real normed vector space $X$ iswith the operator normTwo normed spaces are isometrically isomorphic when there is a bijective linear map between them that preserves norms.

For $y=(y_j)\in {\ell }^{\infty }$, defineThe series is absolutely convergent andso ${\Phi }_y\in ({\ell }^1{)}^{\ast }$ and $\Vert {\Phi }_y\Vert \le \Vert y{\Vert }_{\infty }$. For every $\epsilon >0$, choose $j$ with $|y_j|>\Vert y{\Vert }_{\infty }-\epsilon$ and test on $\mathrm{sgn}(y_j)e_j$. This gives the reverse inequality, hence

Conversely, given $\Phi \in ({\ell }^1{)}^{\ast }$, set $y_j=\Phi (e_j)$. Then $|y_j|\le \Vert \Phi \Vert$, so $y\in {\ell }^{\infty }$. For $x\in {\ell }^1$, its finite partial sums converge to $x$ in norm; continuity of $\Phi$ therefore givesThe representing sequence is unique, and the construction is linear and norm preserving. Thus the duality of l1 and l infinity proves