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I would like to show that a continuous functional on a subspace $W$ of a normed space $(V,\|\cdot \|)$ has a unique continuous extension to $V$ iff $W$ is dense in $V$.

I have proved $(\Leftarrow$). But the converse is currently eluding me! Help would be much appreciated!

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up vote 2 down vote accepted

Assume $W$ is not dense in $V$ and denote the closure of $W$ by $X$. Then the quotient space $V/X$ with the quotient norm is a non-zero normed space and therefore there is some non-zero bounded linear functional $S:V/X\to \mathbb{C}$. Combining $S$ with the quotient map $V\to V/X$ gives you a non-zero bounded linear map from $V$ into $\mathbb{C}$ which is zero on $W$. Call this map $S_1$. Assume $T_1:V\to \mathbb{C}$ be an extension of $T:W\to \mathbb{C}$. Then $T_1+S_1$ is another extension of $T$.

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You have proved that dense implies uniqueness. This is the direction I stated I know, and indeed you have proved it in much the same way as I did. Have you realised the other direction? – user58514 Mar 16 '13 at 18:42

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