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I am learning functional analysis and the Hahn-Banach theorem right now, but cannot wrap my head around the idea of norms of bounded linear functionals and dual spaces for some reason. One of the corollaries to the Hahn-Banach theorem is

For every $x_0 \in E$ there exists $f_0 \in E^*$ such that $||f_0|| = ||x_0||$ and $\langle f_0, x_0 \rangle = ||x_0||^2$.

The proof of this according to my book is by the application of another corollary to the Hahn-Banach theorem:

Let $G \subset E$ be a linear subspace. If $g: G \rightarrow \mathbb{R}$ is a continuous linear functional, then there exists $f \in E^*$ that extends $g$ and such that $$ ||f||_{E^*} = \sup_{x \in G, ||x|| \leq 1} |g(x)| = ||g||_{G^*}. $$

According to the book, by setting $G = \mathbb{R} x_0$ and $g(tx_0) = t||x_0||^2$, we will supposedly get that $||g||_{G^*} = ||x_0||$. I do not quite understand the details of this argument, although I think I get the general idea. What is tripping me up is that it seems that if $g(tx_0) = t||x_0||^2$ is a continuous linear functional, then by the above corollary we would get instead that there exists $f \in E^*$ such that $$ ||f|| = ||g||_{G^*} = \sup_{x \in G, ||x|| \leq 1} |g(x)| = \sup_{tx_0 \in \mathbb{R} x_0, ||t x_0|| \leq 1} |t ||x_0||^2 | = \sup_{t \in \mathbb{R}, t||x_0|| \leq 1} |t ||x_0||^2| = | ||x_0||^2 | = ||x_0||^2 $$ rather than $||f|| = ||x_0||$. Can someone show me what I'm missing here?

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If $\lvert t \rvert \lVert x_0 \rVert\le 1$, then $$\lvert t \rvert \lVert x_0\rVert^2\le 1\cdot\lVert x_0\rVert= \lVert x_0\rVert$$ and this is the maximum value it can attain.

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  • $\begingroup$ Ooooh, I see, thanks! $\endgroup$ – august Feb 5 '13 at 4:08

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