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Suppose that $ (A,\Sigma,m) $ is a measure space and $ H $ is a linear functional on $ {L^{\infty}}(A,\Sigma,m) $. If $$ \mathcal{U} := \left\{ u: A \to \mathbb{R} ~ \Bigg| ~ \text{$ u $ is measurable, bounded and $ \int_{A} u ~ d{m} = 1 $} \right\} $$ and there are functions $ u_{1},u_{2} \in \mathcal{U} $ such that $$ H(u_{1}) \leq 0 \quad \text{and} \quad H(u_{2}) \geq 0, $$ then my question is:

Is there a function $ u_{3} \in \mathcal{U} $ such that $ H(u_{3}) = 0 $?

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The general form of intermediate value theorem is really a statement in topology: the image of a connected set under a continuous map is also connected. In particular, if $H$ is continuous on a connected set where it attains values of both signs, then $H$ attains $0$ as well. Nothing to do with linearity. –  user53153 Jan 1 '13 at 18:27

1 Answer 1

For each $ t \in [0,1] $, define $ v_{t} \in \mathcal{U} $ as follows: $$ v_{t} \stackrel{\text{def}}{=} t \cdot u_{1} + (1 - t) \cdot u_{2}. $$ Clearly, each $ v_{t} $ is measurable and bounded, being a linear combination of measurable and bounded functions. Also, \begin{align} \forall t \in [0,1]: \quad \int_{A} v_{t} \,d{m} &= \int_{A} [t \cdot u_{1} + (1 - t) \cdot u_{2}] \,d{m} \\ &= \int_{A} t \cdot u_{1} \,d{m} + \int_{A} (1 - t) \cdot u_{2} \,d{m} \\ &= t \int_{A} u_{1} \,d{m} + (1 - t) \int_{A} u_{2} \,d{m} \\ &= t \cdot 1 + (1 - t) \cdot 1 \\ &= 1. \end{align} Hence, indeed, $ v_{t} \in \mathcal{U} $ for all $ t \in [0,1] $. By the linearity of $ H $, we get $$ \forall t \in [0,1]: \quad H(v_{t}) = H(t \cdot u_{1} + (1 - t) \cdot u_{2}) = t \cdot H(u_{1}) + (1 - t) \cdot H(u_{2}). $$ Notice that $ H(v_{0}) = H(u_{2}) \geq 0 $ and $ H(v_{1}) = H(u_{1}) \leq 0 $. By applying the Intermediate Value Theorem to the continuous function $$ t \longmapsto t \cdot H(u_{1}) + (1 - t) \cdot H(u_{2}) $$ defined on the closed interval $ [0,1] $, we see that there exists a $ t^{*} \in [0,1] $ for which $ H(v_{t^{*}}) = 0 $. We can therefore set $ u_{3} := v_{t^{*}} $.

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