Here's Theorem 4.3-2 in Introductory Functional Analysis With Applications by Erwine Kreyszig:

Let $X$ be a normed space, let $Z$ be a subspace of $X$, and let $f$ be a bounded linear functional defined on $Z$. Then there exists a bounded linear functional $\tilde{f}$ defined on $X$ such that $$\tilde{f}(x) = f(x) \ \mbox{ for all } \ x \in Z, \ \mbox{ and } \ \Vert \tilde{f} \Vert_X = \Vert f \Vert_Z,$$ where $$\Vert f \Vert_Z \colon= \sup \{ \ \frac{\vert f(z) \vert }{\Vert z \Vert} \ \colon \ z \in Z, \ z \neq 0 \ \} \ \mbox{ if } \ Z \neq \{ 0 \}; \ \mbox{ otherwise } \ \Vert f \Vert_Z = 0.$$ And, $$\Vert \tilde{f} \Vert_X \colon= \sup \{ \ \frac{ \vert \tilde{f}(x) \vert }{ \Vert x \Vert } \ \colon \ x \in X, \ x \neq 0 \ \}.$$

Now let $X \colon= \mathbb{R}^3$ with the Euclidean norm, let $a \colon= (\alpha_1, \alpha_2, 0) \in X$, and let $$Z \colon= \{ \ (\xi_1, \xi_2, \xi_3) \in \mathbb{R}^3 \ \colon \ \xi_3 = 0 \ \}.$$ Let $f$ be defined on $Z$ by $$f(z) \colon= \alpha_1 \xi_1 + \alpha_2 \xi_2 \ \mbox{ for all } \ z \colon= (\xi_1, \xi_2, 0) \in Z.$$ Then what are all the possible linear extensions $\tilde{f}$ of $f$ as gauranteed by Theorem 4.3-2 in Kreyszig?

Here, $$\Vert f \Vert = \Vert a \Vert = \sqrt{ \alpha_1^2 + \alpha_2^2}.$$

Of course, one possible extension $\tilde{f}$ is given by $$\tilde{f}(x) \colon= \alpha_1 \xi_1 + \alpha_2 \xi_2 \ \mbox{ for all } \ x \colon= (\xi_1, \xi_2, \xi_3) \in \mathbb{R}^3. $$

Am I right?

If so, then what are other such extensions $\tilde{f}$, if any?


Any bounded linear functional defined on a subspace of a Hilbert space admits a unique norm preserving extension.

The proof is given here.

More on Banach spaces with unique extension property for functionals you can find in this discussion


Your analysis is correct.

Also your $\tilde f$ is the only such extension. By the Riesz representation theorem, every bounded linear functional g on $\mathbb{R}^3$ (with Euclidean norm) is of the form $g(x) = \left<x, y \right>$ for some $y \in \mathbb{R}^3$. If $g$ is defined by $y = (\alpha_1,\alpha_2,\alpha_3)$ for some $\alpha_3 \neq 0$, then $\lVert g\rVert = \sqrt{\alpha_1^2 + \alpha_2^2 +\alpha_3^2 } > \sqrt{\alpha_1^2 + \alpha_2^2 } = \lVert f \rVert $.


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