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The Hahn-Banach Theorem is described as this:

$X$ be a real vector space and $p$ a sublinear functional on $X$. Furthermore, let $f$ be a linear functional which is defined on a subspace $Z$ of $X$ and satisfies

$f(x) \leq p(x)$ for all $x \in Z$.

Then $f$ could be extended on the whole $X$.


Now is the question, I cannot understand the following application of Hahn-Banach theorem:

Let $X$ be a normed space and let $x_{0} \neq 0$ by any element of $X$. Then there exists a bounded linear functional $\tilde{f}$ such that

$||\tilde{f}|| = 1$, $\tilde{f}(x_{0}) = ||x_{0}||$.


The proof is stated as following:

We consider the subspace of $Z$ of $X$ consisting of all elements $x = \alpha x_{0}$ where $\alpha$ is a scalar. On $Z$ we define a linear functional $f$ by

$f(x) = f(\alpha x_{0}) = \alpha ||x_{0}||$.

$f$ is bounded and has norm $||f|| = 1$ because

$|f(x)| = |f(\alpha x_{0})| = |\alpha|||x_{0}|| = ||\alpha x_{0}|| = ||x||$.

Then based on some extension of Hahn-Banach Theorem, $f$ has a linear extension $\tilde{f}$ from $Z$ to $X$ fulfill the condition.


I think I do not fully understand Hahn-Banach theorem. From my understanding, the functional $f$ in the proof is not linear functional.

Let $x = (-1 + 1) x_{0}$, then f(x) = f(0) = ||0|| = 0 instead of $f(-x_{0} + x_{0}) = f(-x_{0}) + f(x_{0}) = ||x_{0}|| + ||x_{0}|| = 2||x_{0}||$.

Where I got wrong?

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Your reasoning is flawed because $-x_0 = \alpha x_0$ with $\alpha = -1$, hence $f(-x_0) = - \| x_0 \|$.

A proof that $f$ is linear: Let $x, y \in Z$ and $s \in \mathbb R$. We want to show that $f(x + s y) = f(x) + s f(y)$. By definition of $Z$ there exist $\alpha_x, \alpha_y \in \mathbb R$ such that $$x = \alpha_x x_0, \; y = \alpha_y x_0 \;\Rightarrow\; x + s y = (\alpha_x + s \alpha_y) x_0.$$ By definition of $f$ we therefore get $$f(x + s y) = (\alpha_x + s \alpha_y) \| x_0 \| = \alpha_x \| x_0 \| + s \alpha_y \| x_0 \| = f(\alpha_x x_0) + s f(\alpha_y x_0) = f(x) + s f(y).$$

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  • $\begingroup$ It is such a stupid mind trap that I have fallen in... Thanks a lot for dragging me out! And sorry for the late reply, on a vocation those days... $\endgroup$ – Shawn Lee Mar 1 '15 at 3:15
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Nothing say that $f(-x_0) = \| -x_0 \|$, $x_0$ is fixed, it's not true for every $x\in X$ (and indeed false for $-x_0$)

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