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Consider two Hilbert spaces $E,F$, where $F$ is densely embedded into $E$, i.e. $F$ is dense in $E$ and there exists a constant $C>0$ such that $$ \|x\|_E\le C\|x\|_F \ \ \textrm{ for all }x\in F. $$ Now, let $(\cdot,\cdot)$ be an inner product on $E$. For every fixed $y\in E$, set $y(x):= (x,y)$ for all $x\in F$. It follows from the Cauchy-Schwarz that $y(x)\in F'$, the dual of $F'$. Under this identification, what does it mean to say that $E$ is dense in $F'$?

Note: Ultimately, I want to prove that $E$ is dense in $F'$ with the given setting, but right now I just want to know what this statement means before I attempt to prove it.

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    This is impossible. You can't embed a Hilbert space $F$ *densely* into another Hilbert space $E$. You can map $F$ onto a dense subspace of $E$, but this can't be an embedding. You can embed $F$ into $E$, but the range cannot be dense (unless the range is $E$ itself). – 2017-01-14
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    I don't see the problem there, could you clarify? – 2017-01-14

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First we have to clarify what norm we put on $F'$. In general the dual of a normed vector space $X$ is endowed with the norm $$ \Vert f \Vert_{X'} = \sup_{\Vert x \Vert_X =1} f(x). $$

Using the map you describe above, for each $y \in E$ you define $f_y \in F'$ via $f_y(x) = \langle y,x\rangle$. Then $$ \Vert f_y \Vert_{F'} = \sup_{\Vert x \Vert_F =1} f_y(x)= \sup_{\Vert x \Vert_F =1} \langle y,x\rangle. $$ Now to prove what you're after you want to show that for any $f \in F'$ and $\epsilon >0$ there exists $y \in E$ such that $\Vert f - f_y \Vert_{F'} < \epsilon$, i.e. $$ \Vert f- f_y \Vert_{F'} = \sup_{\Vert x \Vert_F =1}( F(x)- \langle y,x\rangle) < \epsilon. $$

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    Shouldn't it be a sequence of $y_n\in E$? Also, how would you characterise $F(x)$ in this case? – 2017-01-14
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    I've used an equivalent notion of density. If you prefer to use the sequential version, then you use the above with $\epsilon = 1/n$. – 2017-01-15