Mechanism Design: Proving that the expected utility is differentiable Given a direct mechanism, we define a buyer's expected utility $u(\theta)$ conditional on her type being $\theta$ by $u(\theta)=\theta q(\theta)-t(\theta)$, where $q:[\underline{\theta},\bar{\theta}]\to[0,1]$ and $t:[\underline{\theta},\bar{\theta}]\to\mathbb{R}$.

We also define that a direct mechanism is incentive-compatible if truth-telling is optimal for every $\theta\in[\underline{\theta},\bar{\theta}]$, i.e., $$u(\theta)\geq \theta q(\theta')-t(\theta'),\quad\forall\theta,\theta'\in[\underline{\theta},\bar{\theta}].$$

LEMMA: For an incentive-compatible direct mechanism, we want to show that for all $\theta$ that $u$ is differentiable, we have $u'(\theta)=q(\theta)$.

PROOF: Consider any $\theta$ for which $u$ is differentiable. Let $\delta&gt;0$. Then by incentive compatibility, we have the following:



I don't understand how from incentive compatibility it follows the first and second inequality 2.6 and 2.8. How can we use the same $\theta$ in the proof while the definition clearly states $u(\theta)\geq \theta q(\theta')-t(\theta'),\forall\theta,\theta'$?
By definition we have
$$
\begin{align*}
u(\theta) &amp; = \theta q(\theta)-t(\theta) \\
\\
u(\theta + \delta) &amp; = (\theta + \delta) q(\theta + \delta)-t(\theta + \delta)
\end{align*}
$$
By incentive compatibility (where $\theta + \delta$ is the true type, $\theta$ is the false type) we have
$$
u(\theta + \delta) \geq (\theta + \delta) q(\theta)-t(\theta)
$$
Using these (I used square brackets for clearer notation, there is no mathematical function to them)
$$
\begin{align*}
u(\theta + \delta) - u(\theta) &amp; =
\left[(\theta + \delta) q(\theta + \delta)-t(\theta + \delta)\right] -
\left[\theta q(\theta)-t(\theta)\right]  \\
\\
u(\theta + \delta) - u(\theta) &amp; \geq
\left[(\theta + \delta) q(\theta)-t(\theta)\right] -
\left[\theta q(\theta)-t(\theta)\right]
\end{align*}
$$
This also holds if you divide by $\delta &gt; 0$.

Similar argument for 2.8., using true type $\theta - \delta$ instead of $\theta + \delta$.