# Is there an intermediate value theorem for linear functionals?

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

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 $u_{3} := v_{t^{*}}$.