[proofplan] The real case follows immediately from the [Real Hahn-Banach Theorem](/theorems/2627): a semi-norm is positively homogeneous and subadditive, and $g \le |g| \le p$ produces a real-linear extension $f \le p$, after which symmetry of the semi-norm upgrades $f \le p$ to $|f| \le p$. The complex case is reduced to the real case via the Bohnenblust-Sobczyk identity $f(x) = f_1(x) - i f_1(ix)$, which expresses any $\mathbb{C}$-linear functional through its real part. Applying the real Hahn-Banach theorem to $\operatorname{Re}(g)$ yields a real-linear extension $f_1$, and the identity defines a complex-linear extension $f$. The bound $|f(x)| \le p(x)$ is recovered by rotating $x$ by a unit phase $\lambda$ that aligns $f(x)$ with the real axis. [/proofplan] [step:Treat the real case via direct application of the Real Hahn-Banach Theorem] Suppose $\mathbb{F} = \mathbb{R}$. A semi-norm $p$ satisfies $p(\lambda x) = |\lambda| p(x)$ for all $\lambda \in \mathbb{R}$, $x \in X$, and the triangle inequality $p(x + y) \le p(x) + p(y)$. In particular $p$ is positively homogeneous (taking $\lambda \ge 0$) and subadditive. The hypothesis gives $g(y) \le |g(y)| \le p(y)$ for all $y \in Y$, so the [Real Hahn-Banach Theorem](/theorems/2627) — whose hypotheses (positive homogeneity, subadditivity, real subspace, dominated linear $g$) are all met — yields a real-linear extension $f: X \to \mathbb{R}$ of $g$ with \begin{align*} f(x) \le p(x) \qquad \text{for all } x \in X. \end{align*} To upgrade this to $|f(x)| \le p(x)$, evaluate at $-x$: linearity of $f$ gives $f(-x) = -f(x)$, while a semi-norm satisfies $p(-x) = |-1| p(x) = p(x)$. So \begin{align*} -f(x) = f(-x) \le p(-x) = p(x), \end{align*} i.e. $f(x) \ge -p(x)$. Combining, $|f(x)| \le p(x)$ for all $x \in X$. This proves the theorem in the real case. [/step] [step:Reduce the complex case to the real case via the real-part construction] Now suppose $\mathbb{F} = \mathbb{C}$. The vector space $X$ may also be regarded as a real vector space (by restriction of scalars). Every $\mathbb{C}$-linear functional $f: X \to \mathbb{C}$ decomposes as \begin{align*} f(x) = f_1(x) + i f_2(x), \qquad f_1, f_2: X \to \mathbb{R} \text{ are } \mathbb{R}\text{-linear}, \end{align*} where $f_1 = \operatorname{Re}(f)$, $f_2 = \operatorname{Im}(f)$. The constraint of complex linearity gives $f(ix) = i f(x)$, i.e. \begin{align*} f_1(ix) + i f_2(ix) = i\bigl(f_1(x) + i f_2(x)\bigr) = -f_2(x) + i f_1(x). \end{align*} Equating real parts: $f_1(ix) = -f_2(x)$, i.e. $f_2(x) = -f_1(ix)$. Substituting into the decomposition yields the **Bohnenblust-Sobczyk identity** \begin{align*} f(x) = f_1(x) - i f_1(ix), \qquad f_1 = \operatorname{Re}(f). \end{align*} Conversely, given any $\mathbb{R}$-linear $f_1: X \to \mathbb{R}$, the formula $f(x) := f_1(x) - i f_1(ix)$ defines a $\mathbb{C}$-linear functional with real part $f_1$, as we verify in the next step. [guided] A complex-linear functional satisfies the strong constraint $f(ix) = i f(x)$, which couples the real and imaginary parts of $f$. We are looking for a way to apply the *real* Hahn-Banach theorem to the complex problem. The natural strategy is to reduce a complex linear functional to a single real-valued object — its real part — and recover the imaginary part from the linearity constraint. Writing $f = f_1 + i f_2$ with $f_1, f_2$ real-linear, the equation $f(ix) = i f(x)$ becomes \begin{align*} f_1(ix) + i f_2(ix) = i f_1(x) - f_2(x), \end{align*} and equating real parts gives $f_2(x) = -f_1(ix)$. So the imaginary part is determined by the real part via the action of multiplication by $i$ on the underlying real vector space. Substituting back: \begin{align*} f(x) = f_1(x) - i f_1(ix), \end{align*} the **Bohnenblust-Sobczyk identity**. The crucial observation is that this is a *bijection* between $\mathbb{R}$-linear functionals $X \to \mathbb{R}$ and $\mathbb{C}$-linear functionals $X \to \mathbb{C}$. So we have reduced the existence of a $\mathbb{C}$-linear extension of $g$ to the existence of an $\mathbb{R}$-linear extension of $\operatorname{Re}(g)$. [/guided] [/step] [step:Apply the real case to $\operatorname{Re}(g)$ and define the complex extension via the Bohnenblust-Sobczyk identity] Define \begin{align*} g_1: Y &\to \mathbb{R}, \\ y &\mapsto \operatorname{Re}(g(y)). \end{align*} Since $g$ is $\mathbb{C}$-linear, $g_1$ is $\mathbb{R}$-linear (regarding $Y$ as a real subspace of $X$). Moreover \begin{align*} |g_1(y)| = |\operatorname{Re}(g(y))| \le |g(y)| \le p(y) \qquad \text{for all } y \in Y. \end{align*} By the real case (proved in the first step), there exists an $\mathbb{R}$-linear extension $f_1: X \to \mathbb{R}$ of $g_1$ with $|f_1(x)| \le p(x)$ for all $x \in X$. Define \begin{align*} f: X &\to \mathbb{C}, \\ x &\mapsto f_1(x) - i f_1(ix). \end{align*} We verify three properties. **$f$ is $\mathbb{C}$-linear.** Real linearity in $x$ is inherited from $\mathbb{R}$-linearity of $f_1$. For multiplication by $i$: \begin{align*} f(ix) = f_1(ix) - i f_1(i \cdot ix) = f_1(ix) - i f_1(-x) = f_1(ix) + i f_1(x) = i\bigl(f_1(x) - i f_1(ix)\bigr) = i f(x), \end{align*} using $\mathbb{R}$-linearity of $f_1$ to write $f_1(-x) = -f_1(x)$. Combined with real linearity, $f$ is $\mathbb{C}$-linear. **$f|_Y = g$.** For $y \in Y$, the Bohnenblust-Sobczyk identity applied to $g$ itself gives $g(y) = \operatorname{Re}(g(y)) - i \operatorname{Re}(g(iy)) = g_1(y) - i g_1(iy)$. Since $f_1|_Y = g_1$ and $iy \in Y$ (as $Y$ is a $\mathbb{C}$-subspace, hence stable under multiplication by $i$), we have $f_1(iy) = g_1(iy)$. Therefore \begin{align*} f(y) = f_1(y) - i f_1(iy) = g_1(y) - i g_1(iy) = g(y). \end{align*} [/step] [step:Bound $|f(x)|$ by rotating to the real axis] Fix $x \in X$. We prove $|f(x)| \le p(x)$. If $f(x) = 0$, the bound is immediate. Otherwise pick a unit-modulus complex number $\lambda \in \mathbb{C}$ with $|\lambda| = 1$ and $\lambda f(x) = |f(x)|$ (concretely $\lambda = \overline{f(x)}/|f(x)|$). Then \begin{align*} |f(x)| = \lambda f(x) = f(\lambda x), \end{align*} where the last equality uses $\mathbb{C}$-linearity of $f$. Now $|f(x)|$ is a real number, so taking real parts of both sides: \begin{align*} |f(x)| = \operatorname{Re}\bigl(f(\lambda x)\bigr) = f_1(\lambda x). \end{align*} By the bound $|f_1| \le p$ on $X$ (from the real case applied earlier): \begin{align*} |f(x)| = f_1(\lambda x) \le |f_1(\lambda x)| \le p(\lambda x). \end{align*} Finally, since $p$ is a semi-norm and $|\lambda| = 1$, $p(\lambda x) = |\lambda| p(x) = p(x)$. Combining: \begin{align*} |f(x)| \le p(x). \end{align*} [guided] The bound $|f_1(x)| \le p(x)$ from the real case is a *real* bound, but we need a *complex* modulus bound on $f$. The way to convert one to the other is the **rotation trick**: any complex number $w \ne 0$ can be rotated to its modulus by multiplying by $\lambda := \overline{w}/|w|$, which has $|\lambda| = 1$ and $\lambda w = |w|$. Apply this to $w = f(x)$: choose $\lambda$ with $|\lambda| = 1$ and $\lambda f(x) = |f(x)|$. By $\mathbb{C}$-linearity, $f(\lambda x) = \lambda f(x) = |f(x)| \in \mathbb{R}_{\ge 0}$. So $f(\lambda x)$ has zero imaginary part, and \begin{align*} |f(x)| = f(\lambda x) = \operatorname{Re}(f(\lambda x)) = f_1(\lambda x). \end{align*} Now we apply the real bound to $\lambda x$: \begin{align*} |f(x)| = f_1(\lambda x) \le |f_1(\lambda x)| \le p(\lambda x) = |\lambda| p(x) = p(x), \end{align*} using that $p$ is a semi-norm and hence absolutely homogeneous. The final step requires the full semi-norm hypothesis $p(\lambda x) = |\lambda| p(x)$ for *complex* scalars; positive homogeneity alone (which is all we used in the real case) would not suffice here. [/guided] [/step] [step:Assemble the conclusion] In the real case (first step) we constructed a real-linear extension $f$ of $g$ with $|f(x)| \le p(x)$. In the complex case (third and fourth steps), $f: X \to \mathbb{C}$ is $\mathbb{C}$-linear, $f|_Y = g$, and $|f(x)| \le p(x)$ for all $x \in X$. In both cases $f$ is the dominated linear extension claimed by the theorem. [/step]