Hahn-Banach and seminorms

Question

Let $V$ be a vector space, $\phi \in V'$, and let $p_1,\dots,p_n$ be seminorms on $V$ s.t. $|\phi (x) | \leq \sum_{k=1}^{n} p_k(x)$ for all $x \in V$. Prove that there are $\phi_1, \dots \phi_n \in V'$ s.t

$$ \phi = \sum_{k=1}^{n} \phi_k, \\ |\phi_k (x) | \leq p_k (x) \ \forall x \in V. $$

I thought about considering the product space $V^n = V \times \dots \times V$, but I'm stuck at trying to figure out a seminorm on $V^n$. If I can figure out a seminorm, I can then consider the subspace $\{(x,\dots,x): x \in V \}$ and the functional $\phi(x,\dots,x) = \psi(x)$ and apply Hahn-Banach theorem.

Answer 1

As you said, considering the subspace $V_0 := \{(x,\dots,x) \colon x \in V \}$ together with a linear functional $g_0(x,\dots,x) := \phi(x)$ is a good point to start.

You have that $$ |g_0(x,\dots,x)| = |\phi(x)| \leq \sum_{k = 1}^n p_k(x) =: p(x,\dots,x) $$ Of course $p(x_1,x_2,\dots,x_n)$ is a seminorm on $V^n$.

The algebraic version of Hahn-Banach gives you an extension $g$ to $V^n$ which is still dominated by $p$, i.e. $$ |g(x_1,\dots,x_n)| \leq p(x_1,\dots,x_n). $$

Let $\phi_i(x) := g(0,\dots,0,\overset{\substack{i\\\smile}}{x},0,\dots,0)$. Then, $$ \sum_i \phi_i(x) = g(x,0,\dots) + g(0,x,0,\dots) + \dots + g(0,\dots,0,x) \\= g(x,\dots,x) = g_0(x,\dots,x) = \phi(x). $$ Furthermore, $$ |\phi_i(x)| = |g(0,\dots,0,\overset{\substack{i\\\smile}}{x},0,\dots,0)| \\\leq p(0,\dots,0,\overset{\substack{i\\\smile}}{x},0,\dots,0) = p_i(x). $$