Trouble connecting pieces of proof in Kesten's seminal paper on Sinai's random walk In Kesten's 1986 paper (Limit distribution of Sinai's Random Walk) we read:

The proof of this lemma uses the fact that the symmetric simple random walk when properly rescaled converges to the Brownian motion. 
So far so good,
But the problem begins when the author says:

We can therefore evaluate $(2.6)$ as $$\lim_{n\to\infty}\frac{d}{d\lambda}E\left\{\exp\left(-\frac\theta n\tilde{s}_n\right);\tilde{M}_n\leqslant y\sqrt{n}\,\text{ and }\,\tilde{m}(\tilde{T}_n)\geqslant-\lambda\sqrt{n}|S_0=0\right\}.\tag{2.7}$$

The convergence in law would give us that
$$\frac{d}{d\lambda} \Bbb{E}\{ e^{-\theta s_+}; M_+ \leq y ; W(s_+)\geq -\lambda\}\\ = \frac{d}{d\lambda} \lim_n \Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -\lambda\sqrt{n}\} $$
We need an argument to take change the derivative with the limit. What would it be?
Accepting that and moving forward, one reads, 

The question then is, how to use (2.11) in (2.7) to obtain Lemma (2.6). Note that there is a $\frac{d}{d\lambda}$ missing in (2.11) (or in the right hand side of (2.6)) and there is a missing $\sqrt{n}$ on (2.7). 
How can we close the proof?
Edit
To prove that
$$\frac{d}{d\lambda} \Bbb{E}\{ e^{-\theta s_+}; M_+ \leq y ; W(s_+)\geq -\lambda\}\\ = \frac{d}{d\lambda} \lim_n \Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -\lambda\sqrt{n}\}\\
= \lim_n\frac{d}{d\lambda}  \Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -\lambda\sqrt{n}\}$$ 
It requires a refined argument. Indeed, let $a(n,h) = 1_{n>1/h}$, then 
$$\lim_{h\to 0} \lim_{n\to \infty} E[a(n,h)] = 1 \\
 \lim_{n\to \infty} \lim_{h\to 0}E[a(n,h)] = 0 \\$$
So we need something more than just a simple Dominated convergence theorem.
We write
$$ \frac{d}{d\lambda} \Bbb{E}\{ e^{-\theta s_+}; M_+ \leq y ; W(s_+)\geq -\lambda\}\\ = \frac{d}{d\lambda} \lim_n \Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -\lambda\sqrt{n}\}\\
\lim_{h\to 0} \frac{1}{h}\big[\lim_n \Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -(\lambda+h)\sqrt{n}\}-\Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -(\lambda)\sqrt{n}\}\big]\\
\lim_{h\to 0} \frac{1}{h}\lim_n \Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; -(\lambda)\sqrt{n}>\tilde m( \tilde T_n)\geq -(\lambda+h)\sqrt{n}\}\\
$$
On the other hand
$$\lim_n \lim_{h\to 0} \frac{1}{h}\Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; -(\lambda)\sqrt{n}>\tilde m( \tilde T_n)\geq -(\lambda+h)\sqrt{n}\}\\=
\lim_n \Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ;\tilde m( \tilde T_n) = -(\lambda)\sqrt{n}\}\\=
\Bbb{E}\{ e^{-\theta  s_+};  M_+ \leq y ;W( s_+) = -(\lambda)\} $$
**Edit 2 **
It may be usefull to add a couple of definitions given by the author:
Here $S_n$ is a symmetric simple random walk tarting from $0$. That is
$S_n = \sum_{i=1}^n X_i$ where $X_i$ are iid. and $P(X_1 = 1) = P(X_1 = -1) = 1/2$

 A: It suffices to differentiate the expectation, you will see that the left-hand side of (2.11) is in fact the derivative of (2.7).
In details:
$$    \frac{d}{d\lambda} \Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -\lambda\sqrt{n}\}  \\
= \lim_{h \to 0}\frac{1}{h}\bigg(\Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -(\lambda + h)\sqrt{n}\} -\Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)\geq -\lambda\sqrt{n}\}\bigg)\\
=\sqrt{n}\lim_{h \to 0}\frac{1}{\sqrt{n}h}\bigg(\Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; 1_{\tilde m( \tilde T_n)\geq -(\lambda + h)\sqrt{n}\}} - 1_{\tilde m( \tilde T_n)\geq -\lambda\sqrt{n}}\}\bigg)  $$
So we need to calculate
$$\lim_{h\to 0+} (\sqrt{n}h)^{-1}\bigg(\Bbb{1}_{\tilde m( \tilde T_n)\geq -(\lambda + h)\sqrt{n}\}} - 1_{\tilde m( \tilde T_n)\geq -\lambda\sqrt{n}}\bigg)\\
 = \lim_{h\to 0+} (\sqrt{n}h)^{-1}\bigg(\Bbb{1}_{\{-\lambda \sqrt{n}\geq\tilde m( \tilde T_n)\geq -(\lambda + h)\sqrt{n}\}} \bigg) = 1_{\tilde m( \tilde T_n) = -\lambda \sqrt{n}}$$
So, inside the expectation it becomes (writting $E_n[f] = \int f \, dP_n)$:
$$ \lim_{h\to 0+} \int e^{-\frac{\theta}{n} \tilde s_n} 1_{ \tilde M_n \leq y\sqrt n} \frac{1}{\sqrt{n}h} \bigg(1_{\tilde m( \tilde T_n)\geq -(\lambda + h)\sqrt{n}\}} - 1_{\tilde m( \tilde T_n)\geq -\lambda\sqrt{n}}\bigg) dP_n \\=\Bbb{E}\{ e^{-\frac{\theta}{n} \tilde s_n}; \tilde M_n \leq y\sqrt n ; \tilde m( \tilde T_n)= -[\lambda\sqrt{n}]\}$$
