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I'm trying to extend a real functional $f:V \rightarrow \mathbb{R}$ ($V$ is a complex vector space, boundedness of $f$ not known a priori) to the complex functional $f_{\mathbb{C}}: V \rightarrow \mathbb{C}$.

If I need H-B, then how can I discover an upper bound needed for Hahn-Banach theorem? Particularly, I need to display that $f$ is dominated by some function $p$, i.e. that $f \leq p$ on a subspace of $V$.

How to do this? Does it follow from subspace properties?

mavavilj
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2 Answers2

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If $V$ is a normed space and $f:V\to\mathbb{R}$ is a bounded linear functional, then $$ u(x) = f(x)-if(ix) $$ is a bounded linear functional from $V$ to $\mathbb{C}$ and $$ \lVert f\rVert = \lVert u\rVert. $$

The above result is from proposition 5.5. in Folland's Real Analysis book and does not rely on Hahn-Banach.

Note: Conversely, if $u : V\to\mathbb{R}$ is a linear functional such that $\Re(u) = f$, then $u(x)= f(x)-if(ix)$.

Quoka
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  • What if $f$ is not known to be bounded, like it's in my question. – mavavilj Mar 25 '18 at 13:42
  • If $f$ is unbounded then so is $u$ since $$\lVert{u}\rVert = \sup_{\lVert{x}\rVert=1}\lvert u(x)\rvert \geq \sup_{\lVert{x}\rVert=1}\lvert \Re u(x)\rvert = \sup_{\lVert{x}\rVert=1}\lvert f(x)\rvert = \lVert f \rVert$$ – Quoka Mar 25 '18 at 13:48
  • I don't know whether $V$ is normed either... So just assume that in my question, it's not. – mavavilj Mar 25 '18 at 14:03
  • Could you clarify your question? What exactly are you trying to do? – Quoka Mar 25 '18 at 14:21
  • If one merely knows the existence of a functional on reals, then how to extend it to complex functional. And if one does not require H-B, then why. – mavavilj Mar 25 '18 at 14:24
  • @mavavilj You don't need Hahn-Banach. If $V$ is a complex vector space, the map $u \mapsto \operatorname{Re} u$ is an $\mathbb{R}$-linear bijection between the space of complex linear $u \colon V \to \mathbb{C}$ and the space of real linear $f \colon V \to \mathbb{R}$. If $V$ is a topological vector space, it's easy to see that $u$ is continuous if and only if $\operatorname{Re} u$ is continuous. – Daniel Fischer Mar 25 '18 at 14:41
  • @mavavilj The answer I provided is still true if you drop the assumption that $V$ is normed and $f$ is bounded – Quoka Mar 25 '18 at 19:40
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Maybe I don't understand your question, but what do you mean by "extending real functional to complex functional". Can you be more precise on what you are doing in particular?

One of the versions of Hahn-Banach theorem says that given a real (or complex) continuous functional on vector subspace $W$ of normed vector space $V$, you can extend it in a continuous functional defined on the whole of $V$.

Here continuity refers to boundedness, as more generally continuity of linear operator in linear topological space is equivalent to boundedness. In our case this means that you should be able to find a constant $C>0$ such that for each $x\in W$ one has $\|f(x)\| \leq C \|x\|$.

ters
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  • I refer to extending in the sense of Hahn-Banach. Since how else is one supposed to display that a real functional does extend to a complex functional? – mavavilj Mar 25 '18 at 13:42
  • I thought of it first, but I didn't see a connection with Hahn Banach Theorem. Now I understand that you want to extend a linear functional $V \to \mathbb{R}$ to a $\mathbb{C}$-linear functional $V \to \mathbb{C}$, which is given in the answer of MathUser_NotPrime and it doesn't rely on Hahn Banach theorem. As I explained in my answer Hahn-Banach allows you to extend a bounded functional from a subspace $W$ to a whole space $V$ and what you have apparently looked for is that every real functional is a real part of some complex functional, which has nothing to do with boundedness. – ters Mar 25 '18 at 14:02
  • But does it have to do with the Hahn-Banach? – mavavilj Mar 25 '18 at 14:04
  • I don't think so. Hahn-Banach allows you to extend linear bounded functional defined on subspace of some vector space to the linear functional defined on whole space.

    Remark that this isn't the same thing as your situation as you have already $V \to \mathbb{R}$ defined (so no subspace in the game) and you want to create out of it $V \to \mathbb{C}$, which is always possible as shown in the answer of MathUser_NotPrime. So no need to invoke big theorems.

     – ters Mar 25 '18 at 14:11
  • I didn't see Hahn-Banach requiring boundedness of the functional "a priori", but rather after considering a functional one must show that there's a $p$ that dominates it. https://en.wikipedia.org/wiki/Hahn-Banach_theorem#Formulation Essentially then it becomes bounded, but it's not known before showing $p$. So my question is about, how to show $p$, if one doesn't know it yet. – mavavilj Mar 25 '18 at 14:16
  • But perhaps it's non-explicitly written that in order to find $p$ it means that the functional must also be bounded. – mavavilj Mar 25 '18 at 14:17
  • Yeah, indeed, what I have written is merely a corollary of more general Hahn-Banach theorem. See for instance en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem where you can find a formulation from Rudin's book which is I believe the one you have with you. But still, you don't have any subspaces in your question, you want to extend a field $\mathbb{R}$ which is in your codomain. Really, I don't think this is something where you want use Hahn-Banach theorem. Read carefully once again hypotheses in the theorem and compare it with your question! – ters Mar 25 '18 at 14:24