Is this inequality valid, or did the authors make a mistake? My question relates to Boucheron et al. (1999). I paraphrase from their proof of Theorem 6 (p.15-16).

Let $\phi(u) = e^u - u - 1$, then for all $\lambda\in \mathbb{R}$ the inequality,
\begin{align}
(1 - e^{-\lambda})\Psi^{\prime}(\lambda) - \Psi(\lambda) \leq v \phi(-\lambda), \qquad (1) 
\end{align}
holds true. Now, considering equality in $(1)$, we obtain a ordinary differential equation,
\begin{align}
(1 - e^{-\lambda})\Psi^{\prime}(\lambda) - \Psi(\lambda) = v \phi(-\lambda), \qquad (2)
\end{align}
which has as a solution $\Phi_0 = v\phi(\lambda)$. We want to show that $\Psi \leq \Psi_0$. In fact if $\Psi_1 = \Psi - \Psi_0$, inequality (1) can be written as,
\begin{align}
(1 - e^{-\lambda})\Psi_{1}^{\prime} - \Psi_1(\lambda) \leq 0 \qquad(3)
\end{align}

That $\Psi_0(\lambda) = v\phi(\lambda)$ is a solution to the posed ODE, is, I believe, clear. By noting that,
\begin{align}
\Psi_1(\lambda) &= \Psi(\lambda) - v(e^{\lambda} - \lambda - 1), \\
\Psi_{1}^{\prime}(\lambda) &= \Psi^{\prime}(\lambda) - v(e^{\lambda} - 1),
\end{align}
inequality $(3)$ follows from $(1)$, since
\begin{align}
(1 - e^{-\lambda})\Psi^{\prime}(\lambda) - \Psi(\lambda) - v (e^{-\lambda} + \lambda - 1) &= (1 - e^{-\lambda})\Psi^{\prime}(\lambda) - \Psi(\lambda) + v(e^{\lambda} - \lambda - 1) - v(e^{-\lambda} + e^{\lambda} - 2),  \\[1em]
&= (1 - e^{-\lambda})\Psi^{\prime}(\lambda) - (\Psi(\lambda) - v\phi(\lambda)) - v(e^{\lambda} - 1)(1 - e^{-\lambda}),\\[1em]
&=  (1 - e^{-\lambda})(\Psi^{\prime}(\lambda) - v(e^{\lambda}-1) - (\Psi(\lambda) - \Psi_{0}(\lambda)), \\[1em]
&= (1 - e^{-\lambda})\Psi_{1}^{\prime}(\lambda)- \Psi_1(\lambda).
\end{align}
Continuing with the paper,

Hence defining $f(\lambda) = \ln(e^{\lambda} - 1)$ and $g(\lambda) = e^{-f(\lambda)}\Psi_1(\lambda)$ we have,
\begin{align}
(1 - e^{-\lambda})[f^{\prime}(\lambda)g(\lambda) + g^{\prime}(\lambda)] - g(\lambda) \leq 0, \qquad (4)
\end{align}
which yields since $f^{\prime}(\lambda)(1 - e^{-\lambda}) = 1$
\begin{align}
(1 - e^{-\lambda})g^{\prime}(\lambda) \leq 0. \qquad (5)
\end{align}

Now if I work this out for myself, i.e., substitute the functions $f(\lambda)$ and $g(\lambda)$, I obtain,
\begin{align}
\Psi_1(\lambda) &= e^{f(\lambda)}g(\lambda), \\
\Psi_{1}^{\prime}(\lambda) &= e^{f(\lambda)}f^{\prime}(\lambda)g(\lambda) + g^{\prime}(\lambda)e^{f(\lambda)}.
\end{align}
Plugging this into $(3)$, I obtain,
\begin{align}
(1-e^{-\lambda})e^{f(\lambda)}\Big(f^{\prime}g(\lambda) + g^{\prime}(\lambda)\Big) &- e^{f(\lambda)}g(\lambda) \leq 0 \\[1em]
(1 -e^{-\lambda})(e^{\lambda} - 1)g^{\prime}(\lambda) &\leq 0. \qquad (6)
\end{align}
Now this extra factor is important, since the authors go on and argue that $g(\lambda)$ is non decreasing on $(-\infty,0)$ and non increasing on $(0,\infty)$. This claim makes sense for the inequality $(5)$, but not so much for $(6)$. What am I missing? Is inequality $(5)$ valid? Or did the authors make a mistake?
Edit
Not surprisingly, there are multiple versions of this paper. In Boucheron et al. (1999) - v2 the authors caught their error. The corresponding proof is still a bit 'hand wavy', but I can follow it just fine. Addition is just for full disclosure.
 A: The argument breaks down if $\lambda < 0$. In that case $f(\lambda) = \ln(e^\lambda - 1)$ is not defined.
In fact the statement that (1) implies $\Psi(\lambda) \le \Psi_0(\lambda)$ for all real $\lambda$ is incorrect. The reason is that the factor $1 - e^{-\lambda}$ in front of $\Psi'$ in (1) changes sign at $\lambda = 0$.
So for $\lambda > 0$, the function $\Psi$ is a subsolution, it satisfies
$$ \Psi'(\lambda) \le \frac{\Psi(\lambda) + v \phi(-\lambda)}{1 - e^{-\lambda}}
$$
and for $\lambda< 0$ it is a supersolution, it satisfies the reverse inequality.
Then $\Psi \le \Psi_0$ is expected to hold for $\lambda > 0$, but for $\lambda < 0 $, the inequality $\Psi \ge \Psi_0$ is is likely to hold.
I am not claiming that this follows easily. While $\Psi(0) = \Psi_0(0) = 0$, there is the additional technical difficulty that the ODE becomes singular at $\lambda = 0$, since the denominator $1 - e^{-\lambda}$ vanishes there.
But for now it seems that there is a gap in the proof of Theorem 6 in that paper.
