I am trying to show the theorem below but I am stuck on how to show $\phi$ is surjective.
Theorem: Let $X$ be a normed space and let $Y$ be a proper dense linear subspace of $X$. Then $Y^*\cong X^*$.
And I believe we need this lemma:
Lemma: Let $X$ be a normed space and let $Y$ be a dense linear subspace of $X$. Let $f\in Y^*$. Then there exists a unique $g\in X^*$ such that $g|_Y=f$ and $\|f\|=\|g\|$.
My attempt is : For all $f\in Y^*$, by the lemma above, there exist a unique $g\in X^*$ such that $g|_Y=f$ and $\|f\|=\|g\|$. Define $\phi:Y^*\rightarrow X^*$ by $\phi(f)=g$. Then $\phi$ is well defined since for all $f\in Y^*$, there exist a unique $g\in X^*$ such that $\phi(f)=g$. To show $Y^*\cong X^*$, we only need to show that $\phi:Y^*\rightarrow X^*$ is a surjective linear isometry. Let $f_1,f_2\in Y^*$ and $\alpha\in\mathbb{F}$. Then $f_1+\alpha f_2\in Y^*$ since $Y$ is a linear subspace of $X$. By the lemma above, there exist $g_1,g_2\in X^*$ such that $g_1|_Y=f_1$, $g_2|_Y=f_2$ and $(g_1+\alpha g_2)|_Y=f_1+\alpha f_2$. We have $\phi(f_1+\alpha f_2)=g_1+\alpha g_2=\phi(f_1)+\alpha \phi(f_2)$. Hence, $\phi$ is linear. By the lemma above, there exist a unique $g\in X^*$ such that $\|f\|=\|g\|$. Hence, $\phi$ is an isometry.
How to show $\phi$ is surjective and is my proof of the rest is ok?
Thank you in advance!!
Surjectivity. Consider $g\in X^*$, and $f=g_{\mid Y}, f\in Y^*$, there exists a unique $h\in X^*$ such that $h_{\mid Y}=f=g_{\mid Y}$, this implies that $h=g$ and $h=\phi(f)=g$.