Understanding the proof of a theorem using Hahn-Banach Theorem.

analysisfunctional-analysishahn-banach-theoremnormed-spacesproof-explanation

Here is the theorem and its proof:

THEOREM 4. Let $X$ be a normed linear space and $x_0$ a nonzero vector in $X$. There is $T\in X^*$ such that $Tx_0=\|x_0\|$ and $\|T\|=1$.

And here is the statement of the Hahn-Banach Theorem we are using:

THEOREM 3. The Hahn-Banach Theorem. Let $X$ be a normed linear space, let $Y\subset X$ be a linear subspace, and let $T$ be a bounded linear functional on $Y$. There there exists a linear functional $\hat T\in X^*$ such that $$\hat Tx=Tx,\ \ \text{ for } x\in Y,\ \ \ \ \text{and } \ \ \ \|\hat T\|_{X^*}=\|T\|_{Y^*}$$

And here are my questions:

  1. Why we defined $T_{0}$ like that? what is the importance of the $\lambda$ in the definition?

  2. How is the norm of $T_{0}$ is 1?

  3. where is the extension of $T_{0}$ that we should have found according to the Hahn-Banach Theorem ? and how the result follows from the Hahn-Banach Theorem ? where are the details for this?

Could anyone help me in answering those questions please?

Best Answer

If $x_0 \neq 0$, you have a subspace $Y= \operatorname{span}_{x_0} = \{ \lambda x_0 : \lambda \in \mathbb{K} \}.$ So every element in this space is a multiple of $x_0$. If you want to define a functional $T_0$ on $Y$, you have to explain what it does on $Y$, and if you write $T_0 (\lambda x_0) = \lambda \| x_0 \|$, you have defined $T_0$ on all of $Y$. Note that this explains the role of $\lambda$ here.

The importance of $T_0$ is the following. Suppose for the moment that $T_0$ is a bounded linear functional on $Y$ with norm $1$, then by the Hahn-Banach theorem you have an extension $T: X \longrightarrow \mathbb{K}$ such that $T\vert_Y = T_0$ and $\| T \| = \| T_0 \|$. This explicitly means $\| T \| = 1$ and $T x_0 = \| x_0 \|$ by the definition of $T_0$ (!), hence your theorem $4$ is proved.

So what we only have to verify are the properties of $T_0$. I leave the linearity to you, this is straightforward. We finally have to show $\| T_0 \| = 1$. Recall therefore the definition of the operator norm, $\| T_0 \| = \sup_{x \neq 0 } | T_0(x) | / \| x \|$. But $x = \lambda x_0$ for a $\lambda \in \mathbb{C}$ and $T_0(x) = \lambda \| x_0 \|$, hence the $\sup$ becomes $1$.

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