[proofplan]
Both statements reduce to the [Semi-Norm Hahn-Banach Theorem](/theorems/2628) by exhibiting a suitable semi-norm $p$ that dominates the data.
For (i), the dominating semi-norm is $p(x) = \|g\|_{Y^*}\|x\|_X$, which is genuinely a semi-norm since $\|\cdot\|_X$ is a norm and $\|g\|_{Y^*} \ge 0$ is a constant. The bound $|g(y)| \le \|g\|_{Y^*}\|y\|_X = p(y)$ holds by definition of the operator norm, so the semi-norm Hahn-Banach theorem applies and yields $f$ with $\|f\|_{X^*} \le \|g\|_{Y^*}$. Equality follows because $f$ extends $g$, so the supremum defining $\|f\|_{X^*}$ already includes the supremum defining $\|g\|_{Y^*}$.
For (ii), we apply (i) to the one-dimensional subspace $Y = \operatorname{span}\{x_0\}$ with the explicit functional $g(\lambda x_0) := \lambda\|x_0\|$, whose norm on $Y$ equals $1$.
[/proofplan]
[step:Prove (i) by applying the semi-norm Hahn-Banach theorem with $p = \|g\|_{Y^*}\|\cdot\|_X$]
Define
\begin{align*}
p: X &\to \mathbb{R}, \\
x &\mapsto \|g\|_{Y^*}\,\|x\|_X.
\end{align*}
**$p$ is a semi-norm.** It is non-negative, satisfies $p(\lambda x) = |\lambda|\, p(x)$ (since $\|\lambda x\|_X = |\lambda|\|x\|_X$), and obeys the triangle inequality $p(x + y) \le p(x) + p(y)$ (since $\|\cdot\|_X$ does and $\|g\|_{Y^*} \ge 0$). Hence $p$ is a semi-norm on $X$.
**Domination of $g$ by $p$ on $Y$.** By definition of the operator norm,
\begin{align*}
|g(y)| \le \|g\|_{Y^*}\,\|y\|_X = p(y) \qquad \text{for all } y \in Y.
\end{align*}
The hypotheses of the [Semi-Norm Hahn-Banach Theorem](/theorems/2628) are now met: $X$ is a real/complex vector space, $p$ is a semi-norm on $X$, $Y \subseteq X$ is a subspace, and $g: Y \to \mathbb{F}$ is linear with $|g| \le p$ on $Y$. Theorem 2628 yields a linear extension $f: X \to \mathbb{F}$ of $g$ with
\begin{align*}
|f(x)| \le p(x) = \|g\|_{Y^*}\,\|x\|_X \qquad \text{for all } x \in X.
\end{align*}
Taking the supremum over $x$ in the closed unit ball $\overline{B}_X = \{x \in X : \|x\|_X \le 1\}$:
\begin{align*}
\|f\|_{X^*} = \sup_{\|x\|_X \le 1} |f(x)| \le \|g\|_{Y^*}.
\end{align*}
[/step]
[step:Verify the reverse inequality $\|f\|_{X^*} \ge \|g\|_{Y^*}$]
Since $f|_Y = g$, for every $y \in Y$ with $\|y\|_X \le 1$,
\begin{align*}
|g(y)| = |f(y)| \le \sup_{\|x\|_X \le 1, x \in X} |f(x)| = \|f\|_{X^*}.
\end{align*}
Taking the supremum over $y \in Y$ with $\|y\|_X \le 1$:
\begin{align*}
\|g\|_{Y^*} = \sup_{y \in Y,\ \|y\|_X \le 1} |g(y)| \le \|f\|_{X^*}.
\end{align*}
Combining with the previous step yields $\|f\|_{X^*} = \|g\|_{Y^*}$, completing the proof of (i).
In particular, $f \in X^*$ since it is a bounded linear functional on $X$.
[/step]
[step:Prove (ii) by applying (i) to the one-dimensional subspace $\operatorname{span}\{x_0\}$]
Fix $x_0 \in X \setminus \{0\}$ and define
\begin{align*}
Y := \operatorname{span}\{x_0\} = \{\lambda x_0 : \lambda \in \mathbb{F}\} \subseteq X.
\end{align*}
Since $x_0 \ne 0$, every element of $Y$ has a unique representation $\lambda x_0$. Define
\begin{align*}
g: Y &\to \mathbb{F}, \\
\lambda x_0 &\mapsto \lambda\,\|x_0\|_X.
\end{align*}
**$g$ is linear and bounded.** Linearity is immediate from the formula. For the norm,
\begin{align*}
|g(\lambda x_0)| = |\lambda|\,\|x_0\|_X = \|\lambda x_0\|_X,
\end{align*}
so $|g(y)| = \|y\|_X$ for all $y \in Y$. In particular $|g(y)| \le \|y\|_X$ shows $\|g\|_{Y^*} \le 1$. Conversely, choosing $y = x_0/\|x_0\|_X \in Y$ with $\|y\|_X = 1$ gives $|g(y)| = 1$. Hence $\|g\|_{Y^*} = 1$.
By part (i) just proved, there exists an extension $f \in X^*$ of $g$ with $\|f\|_{X^*} = \|g\|_{Y^*} = 1$. Furthermore,
\begin{align*}
f(x_0) = g(x_0) = 1 \cdot \|x_0\|_X = \|x_0\|_X,
\end{align*}
which is the second required property.
[/step]
[step:Assemble the conclusion]
Part (i): the functional $f$ constructed in the first two steps lies in $X^*$, restricts to $g$ on $Y$, and has $\|f\|_{X^*} = \|g\|_{Y^*}$.
Part (ii): the functional $f$ constructed in the previous step lies in $X^*$ with $\|f\|_{X^*} = 1$ and $f(x_0) = \|x_0\|_X$.
Both conclusions of the theorem are proved.
[/step]
