If $Y \subset X$ is dense subspace, then why we can think $Y^*$ and $X^*$ is equivalent? I saw 
"If $Y \subset X$ is dense subspace, then we can think $Y^*$ and $X^*$ is equivalent. In fact for $f \in X^*$, $f|_Y \in Y^*$ and the map $ f\to f|_Y$ makes them equivalent"
in a book. But I cannot prove this. I think it means that for $T: X^* \to Y^*$, 
(1) Domain of $T$ = $X^*$
(2) Range of $T = Y^*$
(3) $\|f\|_{X^*} = \| Tf \|_{Y^*}$ 
But I could not show any of this things. Would you please help me?
 A: Here is a sketch: 
1). If $f:Y\to \Omega $ is a continuous linear functional, then it extends uniquely to an $\overset{\sim }{f}:X\to \Omega :$ 
uniqueness is clear since a continuous function is determined by its values on a dense subset of the domain. For existence, define $\overset{\sim }{f}(x)=\lim f(x_n)$ for any $x_n\to x$ and prove that $\overset{\sim }{f}$ is well-defined, noting that as $x_n$ is Cauchy in $X$ so $f(x_n)$ is Cauchy in $\Omega$ (because $f$ is $linear$), that $\overset{\sim }{f}|_{Y}=f$ (more or less trivial), and that $\overset{\sim }{f}$ is linear. 
2). $\left \| \overset{\sim }{f} \right \|_{X\to \Omega}=\left \| f \right \|_{Y\to \Omega}$:
The $≥$ is trivial. For the other direction, note that if $x_n\in Y$ converges to $x\in X$ then $\left \| \overset{\sim }{f(x)} \right \|=\lim \left \| f(x_n) \right \|\le \left \| f \right \|\lim \left \| x_n \right \|=\left \| f \right \|\left \| x \right \|$.
3). Show that the map $f\mapsto \overset{\sim }{f}$ is linear and bijective and then 2). implies that it is an isometry.
Remark: this argument works for any normed linear space. That is, $\Omega$ does not have to be the scalar field. 
