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 $$\tilde{f}(x) = f(x) \mbox{ for all } x \in Z, \ \mbox{ and } \ \left\lVert \tilde{f} \right\rVert_X = \lVert f \rVert_Z,$$ where
$$\lVert f \rVert_Z := \begin{cases} \sup \left\{ \frac{\lvert f(z) \rvert }{\lVert z \rVert } \colon z \in Z, z \neq \mathbf{0} \right\} & \mbox{ if } Z \neq \{ \mathbf{0} \}; \\ 0 & \mbox{ otherwise}. \end{cases}$$ And, $$ \left\lVert \tilde{f} \right\rVert_X \colon= \sup \left\{ \, \frac{ \left\lvert \tilde{f}(x) \right\rvert }{ \lVert x \rVert } \colon x \in X, x \neq \mathbf{0} \, \right\}.$$
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?
Best Answer
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