= Solution
For the supporting inequality, assume first that $F$ is convex and fix $z,z_0$. For $0<t\leq1$,
$$
F(z_0+t(z-z_0))\leq(1-t)F(z_0)+tF(z).
$$
Subtract $F(z_0)$, divide by $t$ and let $t\downarrow0$. <Differentiability> gives
$$
\boxed{F(z)\geq F(z_0)+DF(z_0)\cdot(z-z_0).}
$$
Conversely assume this inequality for every pair. Put $w=(1-t)z_0+tz$ and apply it with base point $w$ to both $z_0$ and $z$. Multiply those two inequalities by $1-t$ and $t$ and add. The <gradient> terms cancel, proving $F(w)\leq(1-t)F(z_0)+tF(z)$. Thus the supporting inequality characterizes a differentiable <convex function>.
For the <integral> inequality let $\bar z=|\Omega|^{-1}\int_\Omega u$. Apply the supporting inequality at $\bar z$ to $u(x)$ and integrate. Its affine term has <integral> $DF(\bar z)\cdot(\int_\Omega u-|\Omega|\bar z)=0$, yielding
$$
\boxed{F\left(\frac1{|\Omega|}\int_\Omega u\right)
\leq\frac1{|\Omega|}\int_\Omega F(u).}
$$
This is <Jensen's inequality>. Since $u$ is integrable, the supporting affine lower bound has integrable negative part, so the right side is well-defined as an extended real <integral>. If its positive part is infinite, the inequality remains valid.
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