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?


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 $$f(x + s y) = (\alpha_x + s \alpha_y) \| x_0 \| = \alpha_x \| x_0 \| + s \alpha_y \| x_0 \| = f(\alpha_x x_0) + s f(\alpha_y x_0) = f(x) + s f(y).$$

  • $\begingroup$ 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

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 $-x_0$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.