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!


I think a good way of thinking about this is through orthogonality. So, let's look at the more general idea about Hahn-Banach for Hilbert spaces.

One can prove Hahn-Banach in Hilbert spaces a little easier. Suppose that we have a continuous linear functional $f$ of norm 1 defined on a closed subspace $B$ of $H$, where $H$ is separable (for example). Then we can choose a vector $x$ which is orthogonal to $B$, say, $\|x\| = 1$ and $\langle x, b \rangle = 0$ for every $b \in B$.

To extend $f$ we define $f(b + tx) = f(b) + \alpha t$ for some $\alpha$. Our claim is that we can choose $\alpha$ such that $\|f\| = 1$.

So, suppose $b+tx$ has norm 1. This means that $\langle b+tx, b+tx \rangle = 1$, so expanding and noticing that the cross term cancels (and supposing we're in the real case) we get $\langle b,b \rangle + t^2 = 1$, namely, $\|b\| \leq 1$

Then $f(b+tx) = f(b) + \alpha t \leq \|b\| + \alpha t$. So if $\alpha = 0$ then $f(b + tx) \leq \|b\| \leq \sqrt{\|b\|^2 + t^2} = \|b + t x\|$. So we can extend one dimension (using the orthogonal dimension) and keep going (separability means induction probably suffices.)

The question is why does this not work in Banach and locally convex spaces?

The issue comes down to orthogonality. The key point in this proof that we had that isn't present in Banach spaces is the following: if $\|b+tx\| \leq 1$ then $\|b\| \leq 1$.

However, this isn't true in general. I will give you an artificial example that perhaps helps: suppose that we took $\mathbb{R}^3$, $B = $span$\{e_1,e_2\}$ as the unit basis vectors and for $x_3$ we took $-0.4 (e_1 + e_2) + c e_3$ (where $c$ is chosen so that this has norm 1).

Then if we look at the vector $u = \frac{\sqrt{2}}{2} (e_1 + e_2)$ and look at $u + x_3$ we get that this has norm less than 0.6 here is the computation sadly.

So in this case, if I had any functional $f$ defined on $B$, if I decided to extend it to all of $\mathbb{R}^3$ through setting $\alpha = 0$ the norm could increase: as a very concrete example suppose that $f(u)$ was the maximum we could get, and $f(u) = 1$. Then $\|f\| = 1$, but if $\alpha = 0$ then $f(u+x_3) = f(u) = 1$, but $\|u+x_3\| < 1$. Thus the norm has increased.

I hope this helps: orthogonality makes a lot of things in life easier, Hahn-Banach is just one of them.

  • $\begingroup$ (What I said about Hilbert spaces is replicating the proof in the Banach case: I don't doubt there are faster ways of seeing it through Riesz representation.) $\endgroup$ – James Kilbane Nov 12 '15 at 10:01
  • $\begingroup$ I still don't quite understand the extension on an example, however. I understand the "proof" you gave with Hilbert spaces, but in the example you provided where are you extending from? I assume you're beginning with a function on $\mathbb{R}^{2}$ defined as $(x,y)\mapsto -0.4(x+y).$ But in this case the operator norm of such a function would be 0.4$\sqrt{2},$ would it not? And to extend it to be defined on $\mathbb{R}^{3},$ c = 0 would still result in the same norm, so I don't understand your comment in the second to last sentence. $\endgroup$ – AlmostTrivial Nov 12 '15 at 17:10
  • $\begingroup$ In the example I'm not extending anything, I'm simply showing that the statement $\|b+tx\| \leq 1 \implies \|b\| \leq 1$ isn't necessarily true if we choose $x$ 'badly'. $\endgroup$ – James Kilbane Nov 12 '15 at 21:21
  • $\begingroup$ I'm talking about any functional $f$ in this case, I'll edit it to make it a little clearer (or make it worse). I've added a specific example: I hope it helps. It somewhat fogs the intention of what I'm saying though. Also what I said in the Hilbert space case is genuinely a proof. $\endgroup$ – James Kilbane Nov 12 '15 at 21:24

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