Extending Positive Functionals: Linearity

Question

How does regularity provide linearity?

Given the full Banach space of bounded functions over a suitable set: $$\mathcal{B}:=\{f:\Omega\to \mathbb{C}:\|f\|_\Omega<\infty\}$$ and a linear subspace $\mathcal{C}\leq\mathcal{B}$.

Consider a linear functional on the subspace: $$I:\mathcal{C}\to\mathbb{C}:f\mapsto I(f)$$ that is bounded and positive: $$|I(f)|\leq \|I\|\cdot\|f\|_\Omega$$ $$I(f)\geq 0\quad(f\geq 0)$$

Extend it formally to all bounded functions by splitting these: $$g=g_{\Re_+}+g_{\Re_-}+ig_{\Im_+}+ig_{\Im_-}$$ and then defining: $$I_E(g):=\sup_{f\in\mathcal{C}:f\leq g} I(f)\quad(g\geq 0)$$ (Note that $\sup_{f\leq g}I(f)=\sup_{0\leq f\leq g}I(f)$.)

This gives a sharp estimate for the bound: $$|I_E(g)|=|\sup_{0\leq f\leq g}I(f)|=\sup_{0\leq f\leq g}I(f)\leq\sup_{0\leq f\leq g}\|I\|\cdot\|f\|_\Omega=\|I\|\cdot\|g\|_\Omega\quad(g\geq 0)$$ and therefore: $$|I_E(g)|=\Re_+ e^{-i\phi}I_E(g)=I_E(\Re_+ e^{-i\phi}g)\leq \|I\|\cdot\|\Re_+ e^{-i\phi}g\|_\Omega\leq\|I\|\cdot\|g\|_\Omega\quad(re^{i\phi}:=I_E(g))$$ (Note, linearity has not been used yet in full!)

Moreover it remains positive: $$I_E(g)\geq 0\quad(g\geq 0)$$

Now, it requires additional regularity assumptions in order to maintain linearity...

But which and how to proceed then?

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As an example to see what actually can happen consider the integral of simple functions: $$I:\mathcal{S}([0,1])\to\mathbb{C}:f\mapsto\int_{[0,1]} f\mathrm{d}x$$

and the characteristic functions: $$\chi_{A},:\mathbb{[0,1]}\to\mathbb{C}$$ $$\chi_{A^c}:\mathbb{[0,1]}\to\mathbb{C}$$ for some nonmeasurable set $A\subseteq[0,1]$.

Then the extension gives: $$I_E(\chi_A+\chi_{A^c})=\mu([0,1])$$ $$I_E(\chi_A)+I(\chi_{A^c})=\mu^*(A)+\mu^*(A^c)$$ but $A$ was nonmeasurable so the integral fails to be linear here.

Answer 1

The problem is that it misses some sort of monotone convergence theorem: $$0\leq g_n(x)\uparrow g(x)\implies I_E(g_n)\uparrow I_E(g)$$ Then that would suggest to consider all those functions which are approximable: $$g\in\mathcal{C}_E:\iff\exists f_n\in\mathcal{C}:f_n(x)\to g(x)$$

Answer 2

You have defined $I$ on the nonnegative functions $g$, then extended to arbitrary $f$ by $I(g) = I(g_{\Re_+})- I(-g_{\Re_-})+i I(g_{\Im_+}) - i I(-g_{\Im_-})$ where $g=g_{\Re_+}+g_{\Re_-}+ig_{\Im_+}+ig_{\Im_-}$ and $g_{\Re_+} \ge 0, g_{\Re_-} \le 0, g_{\Im_+} \ge 0, g_{\Im_-} \le 0$. Now suppose $f = g + h$. Why should $I(f) = I(g) + I(h)$?

Well, if all these functions were in $\mathcal C$ this would be true because $I$ is linear on $\mathcal C$. So we hope for an approximation: if for any $\epsilon > 0$ there are $\tilde{f}, \tilde{g},\tilde{h} \in \mathcal C$ such that $\tilde{f} = \tilde{g} + \tilde{h}$ and $|I(f) - I(\tilde{f})|$, $|I(g) - I(\tilde{g})|$, $|I(h) - I(\tilde{h})| < \epsilon$ then $|I(f) - I(g) - I(h)| < 3 \epsilon$, and since $\epsilon$ is arbitrary that means $I(f) - I(g) - I(h) = 0$. Regularity (which you didn't actually state) should allow you to find such $\tilde{f}, \tilde{g}, \tilde{h}$.