Extensions of Linear Functionals with Equal Norms

# Extensions of Linear Functionals with Equal Norms

Recall from The Hahn-Banach Theorem (Complex Version) that if $X$ is a linear space, $Y \subseteq X$ is a subspace, and $p : X \to [0, \infty)$ is a subadditive function such that $p(\lambda x) = |\lambda|p(x)$ for all $\lambda \in \mathbb{C}$ for all $x \in X$ and if $\varphi : Y \to \mathbb{C}$ is a linear functional that is dominated by $p$ on $Y$, that is:

(1)
\begin{align} \quad | \varphi(y) | \leq p(y), \quad \forall y \in Y \end{align}

Then there exists a linear functional $\Phi : X \to \mathbb{C}$ such that:

(2)
\begin{align} \quad \Phi(y) = \varphi(y), \quad \forall y \in Y \end{align}
(3)
\begin{align} \quad \Phi(x) \leq p(x), \quad \forall x \in X \end{align}

In other words, we can find a linear functional extension $\Phi$ of $\varphi$ that is dominated by $p$ on the whole space $X$.

The theorem below is a nice application of the Hahn-Banach theorem. It tells us that given a linear functional $\varphi : Y \to \mathbb{C}$ we can find a linear functional extension $\Phi$ such that $\| \Phi \| = \| \varphi \|$.

 Theorem: Let $X$ be a normed linear space and let $Y \subset X$ be a subspace. If $\varphi \in Y^*$ then there exists a continuous linear functional $\Phi \in X^*$ such that: 1) $\Phi(y) = \varphi(y)$ for all $y \in Y$. 2) $\| \Phi \| = \| \varphi \|$.
• Proof: Let $\varphi : Y \to \mathbb{C}$ be a continuous linear functional. Define $p : X \to [0, \infty)$ for all $x \in X$ by:
(4)
\begin{align} \quad p(x) = \| \varphi \| \| x \| \end{align}
• We will show that $p$ satisfies all of the conditions of the Hahn-Banach theorem. We first show that $p$ is subadditive. Let $x, x' \in X$. Then:
(5)
\begin{align} \quad p(x + x') = \| \varphi \| \| x + x' \| \leq \| \varphi \| (\| x \| + \| x' \|) \leq \| \varphi \| \| x \| + \| \varphi \| \| x' \| = p(x) + p(x') \end{align}
• So $p$ is subadditive. We now show that for all $x \in X$ and for all $\lambda \in \mathbb{C}$ that $p(\lambda x) = |\lambda| p(x)$. We have that:
(6)
\begin{align} \quad p(\lambda x) = \| \varphi \| \| \lambda x\| = | \lambda | \| \varphi \| \| x \| = | \lambda | p(x) \end{align}
• Lastly, we show that $p$ dominates $| \varphi(y) |$ on $Y$. For all $y \in Y$ we have that:
(7)
\begin{align} \quad | \varphi (y) | \leq \| \varphi \| \| y \| = p(y) \end{align}
• Therefore by the Hahn-Banach theorem there exists a linear function $\Phi : X \to \mathbb{C}$ such that:
(8)
\begin{align} \quad \Phi(y) &= \varphi(y) \quad \forall y \in Y \\ \quad |\Phi(x)| & \leq p(x) \quad \forall x \in X \end{align}
• Therefore we have that:
(9)
\begin{align} \quad \| \Phi \| \| x \| = \| \Phi(x) \| \leq p(x) = \| \varphi \| \| x \| \end{align}
• Hence:
(10)
\begin{align} \quad \| \Phi \| \leq \| \varphi \| \quad (*) \end{align}
• For the reverse inequality, observe that since $\Phi_Y = \varphi$ we must have that:
(11)
\begin{align} \quad \| \Phi \| \geq \| \varphi \| \quad (**) \end{align}
• Combining the inequalities at $(*)$ and $(**)$ shows us that $\Phi : X \to \mathbb{C}$ is a linear functional on $X$ with $\Phi (y) = \varphi(y)$ for all $y \in Y$ and:
(12)
\begin{align} \quad \| \Phi \| = \| \varphi \| \quad \blacksquare \end{align}
 Corollary 2: Let $X$ be a normed linear space and let $Y \subseteq X$ be a subspace. Then for each $x_0 \in X$ there is a continuous linear functional $\Phi \in X^*$ such that: 1) $\Phi(x_0) = \| x_0 \|$. 2) $\| \Phi \| = 1$.
• Proof: Let $x_0 \in X$. Define a function $\varphi : \mathrm{span}(x_0) \to \mathbb{C}$ for all $\lambda x_0 \in \mathrm{span} (x_0)$ by:
(13)
\begin{align} \quad \varphi (\lambda x_0) = \lambda \| x_0 \| \end{align}
• It is easy to verify that $\varphi$ is a continuous linear functional on the subspace $\mathrm{span}(x_0)$ of $X$. By Theorem 1 there exists a continuous linear functional $\Phi : X \to \mathbb{C}$ such that:
(14)
\begin{align} \quad \Phi(\lambda x_0) &= \varphi(\lambda x_0) \quad \forall \lambda x_0 \in \mathrm{span}(x_0) \\ \quad \| \Phi \| &= \| \varphi \| \end{align}
• We see that $\Phi(x_0) = \varphi(x_0) = \| x_0 \|$. Also, we have that $\| \varphi \| = 1$ and so $\| \Phi \| = 1$. $\blacksquare$

For example, consider the normed linear space $\mathbb{R}^n$. Consider the linear functional $\varphi : \mathbb{R}^n \to \mathbb{C}$ defined by:

(15)
\begin{align} \quad \varphi (x_1, x_2, ..., x_n) = x_1 \end{align}

That is, for each $(x_1, x_2, ..., x_n) \in \mathbb{R}^n$, $\varphi(x)$ gives us the first component of $x$. Let $e_1 = (1, 0, ..., 0)$. Then $\varphi$ is a continuous linear functional on $\mathbb{R}^n$ for which:

(16)
\begin{align} \quad \varphi(e_1) = 1 = \sqrt{1^2 + 0^2 + ... + 0^2} = \| e_1 \| \end{align}

And also:

(17)
\begin{align} \quad \| \varphi \| = 1 \end{align}