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Please construct a counterexample for the following: $A$ is normed space and $M$ is a dense subspace of $A$, if there is a functional $f$ such that $f(M) = 0$, then $f=0$.

Besides, if $A$ is a Banach space, does the same conclusion hold?

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Are you assuming $f$ to be continuous? Then $f(M) = 0$ implies $f =0$ independently of completeness of $A$ and is very easy to prove. If not, it is simply wrong: show that a non-continuous functional has a dense kernel (in other words: a functional is continuous if and only if its kernel is closed). – t.b. Dec 20 '11 at 9:01
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Consider functional $$ f:c_{00}(\mathbb{N})\to\mathbb{C}:x\mapsto\sum\limits_{n=1}^{\infty}n x(n) $$ where $c_{00}(\mathbb{N})$ is the space of finitely supported sequences with $\sup$ norm. Its kernel is dense in $c_{00}(\mathbb{N})$, however $f\neq 0$.

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