Question

Q. State and prove Hahn Banach theorem.

Answer 1

Hahn-Banach Theorem :

Statement: Let X be a real vector space and p a sub linear functional on X. Furthermore, let f be a linear functional which is defined on a subspace Z of X and satisfies

$$f(x) \leq g(x) \quad \forall \quad x \in z$$

Then f has a linear extension $f(x)$ from Z to X satisfying

$$f(x) \leq f(x), \forall x \in X,$$

that is, $f(x)$ is a linear functional on X, satisfies (1) on X and

$$f(x) \leq f(x) \quad \forall \quad x \in z.$$

Proof: Now we discuss stepwise, we shall prove:

(I) The set E of all linear extensions g of f satisfying $g(x) \leq p(x)$ on their domain D can be partially ordered and Zorn's lemma yields a maximal element $f(x)$ of E.

(II) $f(x)$ is defined on the entire space X.

(III) An auxiliary relation which was used in (b). We start with part

Now we start part (i)

Let E be the set of all linear extensions g of f which satisfy the condition $g(x) \leq p(x) \forall x \in D$,

Clearly, $E \neq \phi$ since $f \in E$. On E we can define a partial ordering by $g(x) \leq h(x)$ meaning h is an extension of g, that is, by definition, $D(h) \supset D(g)$ and $h(x) = g(x)$ for every $x \in D(g)$. For any chain $C \subset E$ we now define $g(x)$ by $g(x) = g(x) g(x)$ if $x \in D(g) g \in C$. $g(x)$ is a linear functional, the domain being

$$D(g) = \sum_{g \subset C} D(g),$$

which is a vector space since C is a chain. The definition of $g(x)$ is unambiguous. Indeed, for an $x \in D(g_1) \cap D(g_2)$ with $g_1, g_2 \in C$

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we have $g_{1}(x) = g_{2}(x)$ since C is a chain,

so that

$$g_{1}(x) \leq g_{2}(x) \text{ or } g_{2}(x) \leq g_{1}(x)$$

$g \leq \frac{1}{8}$ for all $g \in C$.

Hence $g$ is an upper bound of $C$. Since $C \subset E$ was arbitrary, Zorn's lemma thus implies that $E$ has a maximal element $\frac{1}{f}$. By the definition of $E$, this is a linear extension of $f$ which satisfies

$$\frac{1}{f}(x) \leq p(x) \quad \forall x \in D(\frac{1}{f}),$$

Now we start part (ii)

We now show that $D(\frac{1}{f})$, is all of $X$. Suppose that this is false. Then we can choose a

$y_{1} \in X - D(f)$ and consider the subspace $y_{1}$ of $X$ spanned by $D(\frac{1}{f})$ and $y_{1}$. Note that $y_{1} \neq 0$ since $0 \in D(\frac{1}{f}) \quad x \in Y_{1}$

It can be written $x = y + \alpha y_{1}$

This representation is unique.

In fact, $y + \alpha y_{1} = \frac{1}{y} + \beta y_{1}$ with $\frac{1}{f} \in D(\frac{1}{f})$ implies $y - \frac{1}{y} = \alpha y_{1} - \beta y_{1}$

$$y - \frac{1}{y} \in D(f) \text{ when } y_{1} \notin D(\frac{1}{f})$$

, so that the only solution is $y - \frac{1}{y} = 0$ and $\alpha - \beta = 0$

This means uniqueness. A functional $g_{1}$ on $y_{1}$ is defined by

$$g_{1}(y + \alpha y_{1}) = \frac{1}{f}(y) + \alpha c$$

where $c$ is any real constant. It is not difficult to see that $g_{1}$ is linear. Furthermore, for $\alpha = 0$

we have $g_{1}(y) = \frac{1}{f}(y)$. Hence $g_{1}$ is a proper extension of $(\frac{1}{f})$. that is, an extension such that $D(\frac{1}{f})$ is a proper subset of $D(g_{1})$. Consequently, if we can prove that $g_{1} \in E$ by showing that

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$$g_{1}(x) \leq p(x) \quad \forall x \in D(g_{1}),$$

this will contradict the maximalist of $(\dot{f})$, so that $D(\dot{f}) \neq X$ and $D(\dot{f}) = X$ is true.

Now we start part (iii)

Accordingly, we must finally show that $g_1$ with a suitable $c$ in above equations. We consider any $Y$ and $z$ in $D(\dot{f})$. From (2) and (1) we obtain

$$\begin{array}{l} \dot{f}(y) - \dot{f}(z) = \dot{f}(y - z) \leq p(y - z) \\ = p(y + y_1 - y_1 - z) \\ \leq p(y + y_1) + p(-y_1 - z) \end{array}$$

Taking the last term to the left and the term $(\dot{f})$ to the right,

we have

$$- p(-y_1 - z) - \dot{f}(z) \leq p(y + y_1) - \dot{f}(y)$$

where $Y_1$ is fixed. Since $Y$ does not appear on the left and $z$ not on the right, the inequality continues to hold if we take the supremum over $z \in D(\dot{f})$ on the left and the infimum over $y \in D(\dot{f})$ on the right, call it $m_1$

$$m_0 \leq m_1 \text{ and } m_0 \leq c \leq m_1$$

we have from (7) (8a) (8b) -

$$\begin{array}{l} - p(-y_1 - z) - \dot{f}(z) \leq c \\ c \leq p(y_1 + y_1) - \dot{f}(y) \quad z \in D(\dot{f}) \quad y \in D(\dot{f}) \end{array}$$

We prove (6) first for negative $\alpha$ in equation and then for positive $\alpha$.

$$\begin{array}{l} \alpha < 0 \text{ and put } z = \alpha^{-1} y \\ - p(-y_1 - \alpha^{-1} y) - \dot{f}(\alpha^{-1} y) \leq c \end{array}$$

Multiplication by $-\alpha > 0$ gives

$$\alpha p(-y_1 - \alpha^{-1} y) + \dot{f}(y) \leq -\alpha c$$

From this and (5), using $y - \alpha y_1 = x$, we obtain the desired inequality

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$$g_{1}(x) = \dot{f}(y) + \alpha c \leq -\alpha p (-y_{1} - \alpha^{-1} y) = p (\alpha y_{1} + y) = p(x)$$

For $\alpha = 0$ we have $x \in D(\dot{f})$ and nothing to prove. For $\alpha > 0$ we use (8b) with $y$ replaced by $\alpha^{-1}y$ to get

$$c \leq p (y_{1} + \alpha^{-1} y) - \dot{f}(\alpha^{-1} y)$$

Multiplication by $\alpha > 0$ gives

$$\alpha c \leq -\alpha p (y_{1} + \alpha^{-1} y) - \dot{f}(y) = p(x) - \dot{f}(y).$$

From this and (5),

$$g_{1}(x) = \dot{f}(y) + \alpha c \leq p(x)$$

Hence proved

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