# Norm of linear functional; can we take supremum over dense subset?

If $V \subset H$ are Hilbert spaces and $V$ is dense in $H$, is it true that for $f \in H^*$, $$\lVert f \rVert_{H^*} = \sup_{v \in V} \frac{|f(v)|}{\lVert v \rVert_V}?$$ So I mean can we just take the supremum in the definition of the norm of $f$ over the dense subset?

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A complete subspace is a closed subspace. A dense closed subset is the whole space. You should require only that $V$ is a subspace. – JSchlather Nov 29 '12 at 19:25

Yes: let $v_n$ of norm $1$ such that $f(v_n)\geqslant \lVert f\rVert_{H^*}-n^{-1}$. Then let $v'_n\in V$ such that $\lVert v_n-v'_n\rVert\leqslant n^{-1}$. Then $$|f(v'_n)|=|f(v_n)-f(v_n-v'_n)|\geqslant |f(v_n)|-|f(v_n-v'_n)|\geqslant\lVert f\rVert_{H^*}-n^{-1}-\lVert f\rVert_{H^*}n^{-1},$$ which gives $$\lVert f\rVert(1-n^{-1})^{-1}\geqslant\left|f\left(\frac 1{\lVert v'_n\rVert}v'_n\right)\right|\geqslant (1+n^{-1})^{-1}\left(\lVert f\rVert_{H^*}-n^{-1}(1+\lVert f\rVert_{H^*})\right).$$ The result follows.
In what I did, I don't think I used the property of subspace of $V$. – Davide Giraudo Dec 1 '12 at 12:02
what if the norms of $V$ and $H$ are different? – maximumtag Mar 7 '13 at 18:03