# 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:

- 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:

- Also note that for all $y \in Y$ we have that:

- 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 $x \in X$:

- This also shows that $\| F \| \leq \| f \|$. For the reverse inequality, observe that:

- 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:

- Observe that $f \in Y^*$ since for all $y \in Y$ we have that:

- 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:

- 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. |

**Proof:**We will prove corollary 4 by invoking The Closed Graph Theorem.

- 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:

- 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:

- 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:

- 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 $f(l) = 0$. $\blacksquare$