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Let $(E,\mathcal{E},\mu)$ be a measured space with finite measure $\mu$. We denote with $K$ the space of all real valued functions on $E$, which are $\mu$-a.s. equal. This is a vector space. Now I have a function $\kappa:K\to \mathbb{R}$, which satisfies

  • $\kappa(\lambda f)=\lambda \kappa(f),\forall \lambda\ge 0,f\in K$
  • $\kappa(f+g)\le \kappa(f)+\kappa(g),\forall f,g\in K$
  • $\kappa(f+c)=\kappa(f)-c,\forall f\in K,c\in\mathbb{R}$
  • $\kappa(f)\le\kappa(g)$, if $f\ge g$.

Now we define $\eta(f):=\kappa(-f)$. We have $\eta(1)=1$ and $\eta(f)\le 0$ for $f\le 0$.

Now how can I use Hahn-Banach to get a linear Functional $\Lambda:K\to\mathbb{R}$ with $\Lambda(1)=\eta(1)=1$ and $\Lambda(f)\le\eta(f)$ for all $f\in K$?

Clearly $\kappa$ is sublinear and so is $\eta$. I think as the linear subspace I choose the whole space $K$. But then, which is my linear functional from $K$ to the reals?

  • Your definition of $K$ makes no sense. You probably mean that $K$ is the space of all real-valued functions on $E$ modulo as-equality. – Chris Eagle Jul 04 '13 at 08:13
  • Yes exactly, we are taking equivalance classes. Sorry for that –  Jul 04 '13 at 08:16

1 Answers1

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Since you need $\Lambda(1)=1$, your subspace will have to contain $1$. Since you don't need $\Lambda$ to have any particular value at any other point, nothing else needs to be in your subspace. So just use the span of $1$, in other words the constant functions.

Chris Eagle
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  • So you take the subspace as the linear span of the element $1$ and the linear functional maps all the elements of this linear span to $1$? –  Jul 04 '13 at 08:16
  • @hulik: that's not linear. So no. – Chris Eagle Jul 04 '13 at 08:17
  • that was the reason why I asked. So if we choose as linear subspace the span of $1$, which is the linear functional from this span into the reals, which should be smaller than $\eta$? –  Jul 04 '13 at 08:19
  • @hulik: You seriously can't think of a linear functional $\Lambda$ on that space such that $\Lambda(1)=1$? – Chris Eagle Jul 04 '13 at 08:22
  • Maybe I'm overlooking something, but I also need $\Lambda(f)\le \eta(f)$ for all $f$ in the span of $1$. Is it that trivial? –  Jul 04 '13 at 08:25
  • for a linear combination of $1$, i.e. $f:=\sum a_k 1$ I would define $\Lambda(f)=\sum a_k \Lambda(1)$. and define $\Lambda(1):=1$. Is this right? –  Jul 05 '13 at 09:45
  • I really don't want to bother you, but is my idea in the previous comment correct? is this the wanted linear functional ? –  Jul 09 '13 at 08:35