superhulk
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Let $$u(x)=xq(x)-t(x)$$. Incentive compatibility dictates that $$xq(x)-t(x)\geq xq(z)-t(z)$$, when $$x$$ is the observed private value. Using a little algebraic manipulation, it can be shown that,

$$xq(x)-t(x)+zq(z)\geq xq(z)-t(z)+zq(z)$$, or

$$u(x)+zq(z)\geq u(z)+xq(z)$$, or

$$u(x)\geq u(z) +(x-z)q(z)$$. Let this be $$(1)$$.

Similarly, $$zq(z)-t(z)\geq zq(x)-t(x)$$, when $$z$$ is the observed private value. Thus, from here as well, we have $$u(z)\geq u(x) +(z-x)q(x)$$. Let this expression be $$(2)$$.

From $$(1)$$, we have $$\cfrac{u(x)-u(z)}{x-z} \geq q(z)$$. Similarly, from $$(2)$$, we have $$\cfrac{u(x)-u(z)}{x-z} \leq q(x)$$. Now, we have the expression, \begin{align*} q(z) \leq \cfrac{u(x)-u(z)}{x-z} \leq q(x) \\ -(3) \end{align*}

Let $$x=z+\delta$$. We now have fromEDIT As correctly pointed out by TheoreticalEconomist, if $$x > z$$, expression $$(3)$$ tells us that $$q(x)$$ is monotone. Also, \begin{align*} q(z) \leq \cfrac{u(z+\delta)-u(z)}{z + \delta -z} \leq q(z+\delta) \\ -(4) \end{align*}

As as $$\delta \rightarrow 0$$$$u$$ is a convex function, it will follow from $$(4)$$is absolutely continuous. This tells us that $$u'(x)=q(x)$$$$u$$ is differentiable almost everywhere. Thus, wherever $$u$$ is differentiable, we have $$u'(x) = q(x)$$.

superhulk
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• 11

Let $$u(x)=xq(x)-t(x)$$. Incentive compatibility dictates that $$xq(x)-t(x)\geq xq(z)-t(z)$$, when $$x$$ is the observed private value. Using a little algebraic manipulation, it can be shown that,

$$xq(x)-t(x)+zq(z)\geq xq(z)-t(z)+zq(z)$$, or

$$u(x)+zq(z)\geq u(z)+xq(z)$$, or

$$u(x)\geq u(z) +(x-z)q(z)$$. Let this be $$(1)$$.

Similarly, $$zq(z)-t(z)\geq zq(x)-t(x)$$, when $$z$$ is the observed private value. Thus, from here as well, we have $$u(z)\geq u(x) +(z-x)q(x)$$. Let this expression be $$(2)$$.

From $$(1)$$, we have $$\cfrac{u(x)-u(z)}{x-z} \geq q(z)$$. Similarly, from $$(2)$$, we have $$\cfrac{u(x)-u(z)}{x-z} \leq q(x)$$. Now, we have the expression, \begin{align*} q(z) \leq \cfrac{u(x)-u(z)}{x-z} \leq q(x) \\ -(3) \end{align*}

Let $$x=z+\delta$$. We now have from $$(3)$$, \begin{align*} q(z) \leq \cfrac{u(z+\delta)-u(z)}{z + \delta -z} \leq q(z+\delta) \\ -(4) \end{align*}

As $$\delta \rightarrow 0$$, it will follow from $$(4)$$ that $$u'(x)=q(x)$$.