Corollaries to the Hahn-Banach Theorem

# Corollaries to the Hahn-Banach Theorem

Recall from The Hahn-Banach Theorem (Real Version) page that if $X$ is a linear space, $Y \subseteq X$ is a linear subspace, and $\varphi : Y \to \mathbb{R}$ is a linear functional for which there exists a sublinear functional $p : X \to \mathbb{R}$ which dominates $\varphi$, i.e., $\varphi(y) \leq p(y)$ for all $y \in Y$ then there exists a linear function $\Phi : X \to \mathbb{R}$ for which:

• 1. $\Phi(y) = \phi(y)$ for all $y \in Y$ ($\Phi$ is an extension of $\varphi$).
• 2. $\Phi(x) \leq p(x)$ for all $x \in X$ ($\Phi$ is dominated by $p$).

We will now look at some corollaries to the Hahn-Banach theorem.

 Corollary 1: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $Y \subseteq X$ be a linear subspace. If $f \in Y^*$ then there exists an $F \in X^*$ such that $F(y) = f(y)$ for all $y \in Y$ and $\| F \| = \| f \|$.

Corollary 1 says that if $f$ is a bounded linear functional defined on a subspace $Y$ of a normed linear space $X$ then $f$ can be extended to a bounded linear functional $F$ defined on all of $X$ which preserves the norm on $f$.

• Proof: Let $p : X \to \mathbb{R}$ be defined for all $x \in X$ by:
(1)
\begin{align} \quad p(x) = \| f \| \| x \|_X \end{align}
• Note that since $f : Y \to \mathbb{R}$ is a bounded linear operator, $\| f \|$ is well-defined. Note that $p$ is certainly a sublinear functional, since for all $x_1, x_2 \in X$ and for all $\lambda \geq 0$ we have that:
(2)
\begin{align} \quad p(x_1 + x_2) &= \| f \| \| x_1 + x_2 \|_X \leq \| f \|[\| x_1 \|_X + \| x_2 \|_X] = \| f \| \| x_1 \|_X + \| f \| \| x_2 \|_X = p(x_1) + p(x_2) \\ p(\lambda x_1) &= \| f \| \| \lambda x_1\| = \lambda \| f \| \| x_1 \|_X = \lambda p(x_1) \end{align}
• Also note that for all $y \in Y$ we have that:
(3)
\begin{align} \quad f(y) \leq |f(y)| \leq \| f \| \| y \|_X = p(y) \end{align}
• So $f$ is dominated by $p$. So by the Hahn-Banach theorem, there exists a linear functional $F : X \to \mathbb{R}$ such that $F(y) = f(y)$ for all $y \in Y$ and such that $F(x) \leq p(x)$ for all $x \in X$.
• All that remains to show is that $F$ is bounded and $\| F \| = \| f \|$. First, $F$ is bounded since for every $x \in X$:
(4)
\begin{align} \quad |F(x)| \leq p(x) = \| f \| \| x \|_X \end{align}
• This also shows that $\| F \| \leq \| f \|$. For the reverse inequality, observe that:
(5)
\begin{align} \quad \| F \| = \sup_{x \in X, \| x \| = 1} |F(x)| \geq \sup_{y \in Y, \| y \| = 1} |f(y)| = \| f \| \end{align}
• Hence $\| F \| = \| f \|$. $\blacksquare$
 Corollary 2: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $x_0 \in X$. Then there exists an $F \in X^*$ such that $\| F \| = 1$ and $F(x_0) = \| x_0 \|_X$.
• Proof: Let $Y = \{ \lambda x_0 : \lambda \in \mathbb{R} \} = \mathbb{R}x_0$. Then $Y$ is a subspace of $X$. Define a linear functional$f : Y \to \mathbb{R}$ for all $y = \lambda x_0 \in Y$ by:
(6)
\begin{align} \quad f(y) = f(\lambda x_0) = \lambda \| x_0 \|_X \end{align}
• Observe that $f \in Y^*$ since for all $y \in Y$ we have that:
(7)
\begin{align} \quad |f(y)| = |\lambda \| x_0 \||_X = \| \lambda x_0 \| = \| y \| \end{align}
• Which shows us that $\| f \| = 1$. Furthermore, note that $f(x_0) = \| x_0 \|_X$. By Corollary 1, there exists an $F \in X^*$ such that $F(y) = f(y)$ for all $y \in Y$ and $\| F \| = \| f \|$. So $F(x_0) = f(x_0) = \| x_0 \|_X$ and $\| F \| = \| f \| = 1$. $\blacksquare$
 Corollary 3: Let $(X, \| \cdot \|_X)$ be a normed linear space. If for all $f \in X^*$ we have that $f(x) = f(y)$ then $x = y$.
• Proof: Suppose instead that $x \neq y$. Then $x - y \neq 0$. By corollary 2, there exists a linear functional $F \in X^*$ such that $\| F \| = 1$ and $F(x - y) = \| x - y \|_X$. Since $F$ is linear, the second equality implies that:
(8)
\begin{align} \quad F(x) - F(y) = \| x - y \|_X \end{align}
• But $\| x - y \|_X \neq 0$ since $x - y \neq 0$. So $F(x) \neq F(y)$ which is a contradiction since $F \in X^*$. So we must have that $x = y$. $\blacksquare$
 Corollary 4: Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a linear operator. If for every $f \in Y^*$ we have that $f \circ T \in X^*$ then $T$ is a bounded linear operator.
• Let $(x_n)$ be a sequence in $X$ that converges to $x$ and let $(T(x_n))$ converge to $y$. Since $f$ and $f \circ T$ are continuous for all $f \in Y^*$ we have that:
(9)
\begin{align} \quad \lim_{n \to \infty} f(T(x_n)) &= f \left ( T \left ( \lim_{n \to \infty} x_n \right ) \right ) = f(T(x)) \\ \quad \lim_{n \to \infty} f(T(x_n)) &= f \left ( \lim_{n \to \infty} T(x_n) \right ) = f(y) \end{align}
• So $f(T(x)) = f(y)$ for all $f \in Y^*$. By corollary 3, this implies that $T(x) = y$. By the Closed Graph Theorem, $T$ is a bounded linear operator. $\blacksquare$
 Corollary 5: Let $X$ be a normed linear space, let $L \subset S$ by a nonempty linear subspace. If $y \in S$ is such that $d(y, L) = \delta > 0$ then there exists an $F \in X^*$ such that $\| F \| \leq 1$, $f(y) = \delta$, and $f(l) = 0$ for all $l \in L$.

Recall that $d(y, L)$ denotes the distance from the point $y$ to the set $L$ and is given by $d(y, L) = \inf_{l \in L} \| y - l \|$.

• Proof: Let $S = L + \mathbb{R}y = \{ l + \lambda y : l \in L \}$. Then $S$ is a subspace of $X$. Define a function $f : S \to \mathbb{R}$ for all $(l + \lambda y) \in S$ by:
(10)
\begin{align} \quad f(l + \lambda y) = \lambda \delta \end{align}
• Note that $f$ is well-defined, for if $l_1 + \lambda_1 y = l_2 + \lambda_2 y$ then $l_1 - l_2 = (\lambda_2 - \lambda_1)y$ so if $\lambda_1 \neq \lambda_2$ then $y = \frac{l_1 - l_2}{\lambda_2 - \lambda_1}$ which implies that $y \in L$, which is a contradiction since $d(y, L) = \delta > 0$. So $\lambda_1 = \lambda_2$, i.e., $f(l_1 + \lambda_1y) = f(l_2 + \lambda_2y)$.
• Observe that $f$ is linear. $f$ is also bounded since for all $(l + \lambda y) \in S$ we have that:
(11)
\begin{align} \quad |f(l + \lambda y)| = |\lambda \delta| = |\lambda| d(y, L) = |\lambda| \inf_{s \in L} \| s - y \| = \inf_{s \in L} \| s - \lambda y \| \leq \| l + \lambda y \| \end{align}
• Also $\| f \| \leq 1$. By Corollary 1 there exists an $F \in X^*$ such that $F(s) = f(s)$ for all $s \in S$ and $F \| = \| f \|$. So $\| F \| \leq 1$, $F(y) = f(y) = \delta$, and if $l \in L$ then $f(l) = 0$. $\blacksquare$