TITLE: Gronwall's Lemma (Discrete version)
QUESTION [10 upvotes]: I have the following exercise:
Gronwall's lemma (Discrete version):
Let $(u_n)$ and $(w_n)$ be nonnegative sequences satisfying
$$
u_n \leq \alpha + \sum_{k=0}^{n-1}u_kw_k \quad \forall n.
$$
Then for all $n$ it holds
$$
u_n \leq \alpha \exp\biggl( \sum_{k=0}^{n-1} w_k \biggr).
$$
Proof the lemma by the following steps:
(i) Verify the identity
$$
1+\sum_{k=0}^{n-1}\biggl( \prod_{l = 0}^{k-1} (1+w_l) \biggr)w_k \leq \prod_{k = 0}^{n-1} (1+w_k).
$$
(ii) Proof by induction that for all $n$ it holds
$$
u_n \leq \alpha \prod_{k = 0}^{n-1} (1+w_k).
$$
(iii) Deduce the lemma.
I tried to solve this exercise by myself without results on points (ii) and (iii) (I already know how to proof the first point). Can someone provide some ideas or a step-by-step solution?

REPLY [2 votes]: (ii) basis Show that statement holds for $n=1$ using $u_0 \leq \alpha$. inductive step Assume $$u_j \leq \alpha \cdot \prod_{k=0}^{j-1} (1+w_k) \qquad (j=0,\ldots,n) \tag{1} $$ Then $$\begin{align} u_{n+1} &\leq \alpha+ \sum_{j=0}^{n} w_j \cdot u_j \stackrel{(1)}{\leq} \alpha+\sum_{j=0}^{n} \alpha \cdot w_j \cdot \prod_{k=0}^{j-1} (1+w_k) \\ &= \alpha \cdot \left(1+ \sum_{j=0}^n  \prod_{k=0}^{j-1} (1+w_k) \cdot w_j \right) \stackrel{(i),\alpha \geq 0}{\leq} \alpha \cdot \prod_{k=0}^n (1+w_k) \end{align}$$ i.e. $(1)$ holds also for $j=n+1$.
(iii) Hint Note that $1+x \leq e^x$ for all $x \geq 0$.