I am currently enrolled in a functional analysis course and am experiencing some troubles with applying the Hahn-Banach theorem we discussed with regards to extending linear functional. In particular, we were given this practice problem:
Let $L:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be defined $L(x_{1},x_{2}) = 3x_{1} + 4x_{2}.$ Let $C$ be the norm of $L.$ Evaluate $C$, and then construct explicitly an extension $\tilde{L}:\mathbb{R}^{3}\rightarrow\mathbb{R}$ with the same norm.
I am going to present my logic and attempt at the problem, and would appreciate any advice!
Firstly, to determine $|||L|||,$ the operator norm of $L,$ by definition we determine the supremum of $|L(x)|$ over all $x\in\mathbb{R}^{2}$ such that $||x|| = 1$ where $||\ ||$ is the usual Euclidean norm. In other words, all $x$ that lie on the unit circle. I determined that the supremum was 5. This is where I am having some difficulty in applying the Hahn-Banach theorem. It says that if $L(x) \leq p(x)$ for every $x\in\mathbb{R}^{2},$ then we can extend it to a function $\tilde{L}(x + th) = L(x) + \alpha t$ such that $\tilde{L}\leq p(x)$ on now $\mathbb{R}^{3}.$ However, I supposed that $p(x)$ was to be taken as the Euclidean norm, and it is not true that $L(x)\leq ||x||?$ Nonetheless, I attempted to extend $L$ as follows.
To extend $L$ to a linear functional $\tilde{L}$ from $\mathbb{R}^{3},$ we need to have it be \begin{equation*} \tilde{L}(x_{1},x_{2},x_{3}) := L(x + x_{3}t) = L(x) + x_{3}\alpha \end{equation*} where $\alpha = L(t),\ x = (x_{1},x_{2})$ and $x_{3}\in\mathbb{R}.$ Moreover, it must be such that $|||\tilde{L}||| = 5.$ To do this, as the operator norm is take over all $||x|| = 1,$ we must consider $(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}$ such that $x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 1.$ That is, along the sphere of radius 1. With this, to determine the maximum value of $\tilde{L},$ it will be more convenient to convert from Cartesian coordinates into spherical with $\rho = 1$ and $\theta\in [0,2\pi]$ and $\phi\in[0,\pi].$ After this, our linear functional $\tilde{L}$ becomes \begin{equation*} \tilde{L}(\theta,\phi) = \sin{\phi}(3\cos{\theta} + 4\sin{\theta}) + \alpha\cos{\phi} \end{equation*} To determine its maximum value, we must determine when both the partial with respect to $\theta$ and with respect to $\phi$ are both zero to obtain candidates of being a maximum. These partials are, \begin{equation*} \tilde{L}_{\theta} = \sin{\phi}(4\cos{\theta}-3\sin{\theta}),\quad \tilde{L}_{\phi} = \cos{\phi}(3\cos{\theta}+4\sin{\theta})-\alpha\sin{\phi} \end{equation*} The first partial yields $\phi = 0,\pi$ and $\theta = \tan^{-1}{\frac{4}{3}}.$ When considering $\phi = 0,\pi,$ the corresponding $\theta$ the partial with respect to $\phi$ yields $\theta = \tan^{-1}{\left(-\frac{3}{4}\right)}.$ As for $\theta = \tan^{-1}{\frac{4}{3}},$ the corresponding $\phi$ obtained from $\tilde{L}_{\phi}$ is $\phi = \tan^{-1}{\frac{5}{\alpha}}.$ Now, it is then necessary to determine the second-order partials and calculate $D = \tilde{L}_{\theta\theta}\tilde{L}_{\phi\phi} - \tilde{L}_{\theta\phi}^{2}$ for all potential 3 points, but simply plugging in the values yields a clear result that the maximum occurs at the point $\left(\tan^{-1}{\frac{4}{3}},\tan^{-1}{\frac{5}{\alpha}}\right),$ which in turn is \begin{equation*} \tilde{L}\left(\tan^{-1}{\frac{4}{3}},\tan^{-1}{\frac{5}{\alpha}}\right) = \sin{\tan^{-1}{\frac{5}{\alpha}}}\left(3\cos{\tan^{-1}{\frac{4}{3}}} + 4\sin{\tan^{-1}{\frac{4}{3}}}\right) + \alpha\cos{\tan^{-1}{\frac{5}{\alpha}}} \end{equation*} \begin{equation*} = \sqrt{25 + \alpha^{2}} \end{equation*} Thus, we obtain that \begin{equation*} |||\tilde{L}||| = \mathrm{sup}_{||x|| = 1}|\tilde{L}(x)| = \sqrt{25+\alpha^{2}} \end{equation*} but for $|||\tilde{L}|||$ to be equal to 5, we must then require that $\alpha = 0.$
So in other words, the extending linear functional would just be the same $L$ given, but that doesn't seem right. I feel that my doubt particularly stems from my uncertainty in understanding the HB theorem. I would like to say I understand the proof and how to show that such an extension exists, but we did not discuss examples/applications on a problem like this, so I don't know where I stand. Moreover, the professor asked what about extending it from $\mathbb{R}^{3}$ to $\ell^{2}?$ After this first step, however, I would imagine my result would be the same. I would appreciate any help and recommendations to better understand the Hahn-Banach theorem. Thanks in advance!