Oleinik's entropy condition I can't figure out how the proof the implication $\Leftarrow$ from the
following problem:
For $f \in C^2(\mathbb{R})$ and $u_l \neq u_r$ the function
\begin{equation*}
  u(x,t) = \begin{cases}
    u_l &, x < st, \\
    u_r &, x > st,
  \end{cases} \quad s = \frac{f(u_r) - f(u_l)}{u_r - u_l}
\end{equation*}
is a weak solution of the Riemann-Problem
\begin{equation*}
  u_t + f(u)_x = 0, \quad u(x, 0) = \begin{cases}
    u_l &, x < 0, \\
    u_r &, x > 0.
  \end{cases}
\end{equation*}
Show that $u$ satisfies Oleinik's entropy condition if and only if for each convex entropy
$\eta \in C^2(\mathbb{R})$ and corresponding entropy flux $\psi \in C^1(\mathbb{R})$ the inequality
\begin{equation*}
  \eta(u)_t + \psi(u)_x \leq 0
\end{equation*}
holds weakly.
I have proofed that the above inequality holds weakly iff
\begin{equation}
-s(\eta(u_r) - \eta(u_l)) + \psi(u_r) - \psi(u_l) \leq 0
\end{equation}
holds.
So far I tried to rearrange the inequality and use the mean value theorem to show the oleinik entropy condition by contraposition, which
didn't get me quite far.
 A: I assume you mean with Oleinik's entropy condition this inequalities:
$$ \frac{f(v) - f(u_l) }{v - u_l} \geq s \geq \frac{f(v) - f(u_r)}{v -u_r}$$
I present the proof I found in lecture notes on Numerics of Conservation Laws by Siddartha Mishra.
Start with the Fundamental Theorem of Calculus
$$\eta (u_r) - \eta (u_l) = \int_{u_l}^{u_r} \eta'(v) dv$$
Using integration by parts,
$$\int_{u_l}^{u_r} \eta'(v) \cdot 1 dv = \eta'(v)(v - u_l) \Big \vert^{u_r}_{u_l} -  \int_{u_l}^{u_r} \eta''(v) (v - u_l) dv \\
= \eta'(u_r)(u_r - u_l) - \eta'(u_l) \cdot 0 - \int_{u_l}^{u_r} \eta''(v) (v - u_l)$$
For $\psi$, we also use first the Fundamental Theorem of Calculus to obtain
$$\psi(u_r) - \psi(u_l) = \int_{u_l}^{u_r} \psi'(v) dv $$
where $\psi'$ fulfills (if forming an entropy pair with $\eta$)
$$\psi' = \eta' f'.$$
Thus,
$$\int_{u_l}^{u_r} \psi'(v) dv = \int_{u_l}^{u_r} \eta'(v) f'(v) dv $$
Again, use integration by parts to obtain
$$\int_{u_l}^{u_r} \eta'(v) f'(v) dv = \eta'(v) \big(f(v) - f(u_l) \big) \Big \vert^{u_r}_{u_l} - \int_{u_l}^{u_r} \eta''(v) \big(f(v) - f(u_l) \big) d v\\
= \eta'(u_r) \big(f(u_r) - f(u_l) \big) - \eta'(u_l) \cdot 0 - \int_{u_l}^{u_r} \eta''(v) \big(f(v) - f(u_l) \big) d v$$
Plugging this into your result:
$$ \psi(u_r) - \psi(u_l) - s \big( \eta(u_r) - \eta(u_l) \big) = \eta'(u_l) \Big[\big(f(u_r) - f(u_l) \big) - s \big(u_r - u_l\big) \Big] \\
+ \int_{u_l}^{u_r} \eta''(v) \Big[ s \big(v - u_l \big) - \big(f(v) - f(u_l) \big)  \Big] d v \overset{!}{\leq} 0$$
By the Rankine-Hugoniot condition you also stated in the question the first summand on the RHS drops out.
Since $\eta''(v)$ is a strictly convex function we have that $\eta''(v) > 0 \: \forall \: v$.
Thus, we must have
$$ s \big(v - u_l \big) - \big(f(v) - f(u_l) \big) \leq 0 \Leftrightarrow s \leq \frac{f(v) - f(u_l)}{v - u_l} $$
which is exactly the first inequalitiy.
To show the second equality, use the fundamental theorem of calculus with changed boundaries, i.e,
$$\eta (u_r) - \eta (u_l) = - \int_{u_r}^{u_l} \eta'(v) dv$$
which adds the necessary $-$ in front of the integrals to flip the inequality and gets you the $u_r, f(u_r)$.
