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Let $C[0,1]$ with the supremum norm. It's easy to see that $M=\{f\in C[0,1]:f(0)=0\}$ is a closed subspace and so $C[0,1]/M$ is a Banach space. But I'm having trouble in finding a Banach space isometric to $C[0,1]/M$.

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up vote 5 down vote accepted

For a given $a\in[0,1]$ consider linear bounded functional $$ \operatorname{ev}_a:C[0,1]\to\mathbb{C}:f\mapsto f(a) $$ Obviously $M=\operatorname{Ker}(\operatorname{ev}_0)$, i.e. $M$ is a kernel of bounded linear functional so $C([0,1])/M\cong \mathbb{C}$.

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Basically, the first isomorphism theorem. – ncmathsadist May 14 '13 at 13:23

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