How is the Gronwall lemma used in this paper?

Question

Let $(X_t, t \ge 0)$ be a $\mathbb R^d$-valued stochastic process. Let $\lambda>0$. Assume we have $\mathbb E [|X_0|^2] < \infty$ and $$ \mathbb E [|X_t|^2] - \mathbb E [|X_0|^2] \le -2 \lambda \int_0^t \mathbb E [|X_s|^2] \, \mathrm d s + 2t (\lambda |\mathbb E [X_0]|^2 +1) \quad \forall t \ge 0. $$

It is mentioned in the proof of Lemma 5.2 of the paper Convergence to equilibrium for granular media equations and their Euler schemes that

By Gronwall's lemma, $$ \mathbb E [|X_t|^2] \le \bigg [ |\mathbb E [X_0]|^2 + \frac{1}{2\lambda} \bigg ] (1- e^{-2\lambda t}) + \mathbb E [|X_0|^2] e^{-2\lambda t}. $$

Could you explain which form of the Gronwall lemma is used in the paper?

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On the other hand, the form of Gronwall lemma on Wikipedia seems not applicable in this case, i.e.,

Let $I$ denote an interval of the real line of the form $[a, \infty)$ or $[a, b]$ or $[a, b)$ with $a<b$. Let $\alpha, \beta$ and $u$ be realvalued functions defined on $I$. Assume that $\beta$ and $u$ are continuous and that the negative part of $\alpha$ is integrable on every closed and bounded subinterval of $I$.

• (a) If $\beta$ is non-negative and if $u$ satisfies the integral inequality $$ u(t) \leq \alpha(t)+\int_a^t \beta(s) u(s) \mathrm{d} s, \quad \forall t \in I, $$ then $$ u(t) \leq \alpha(t)+\int_a^t \alpha(s) \beta(s) \exp \left(\int_s^t \beta(r) \mathrm{d} r\right) \mathrm{d} s, \quad t \in I. $$
• (b) If, in addition, the function $\alpha$ is non-decreasing, then $$ u(t) \leq \alpha(t) \exp \left(\int_a^t \beta(s) \mathrm{d} s\right), \quad t \in I. $$

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Below is the screenshot I took from the paper.

Answer 1

$\newcommand\al\alpha\newcommand\be\beta\newcommand\la\lambda$The reasoning in the paper is probably as follows:

For real $t\ge0$, letting $$u(t):=2\la(E|X_t|^2-|EX_0|^2)-1,$$ $$\al(t):=-1+2\la(E|X_0|^2-|EX_0|^2),\quad\be(t):=-2\la,$$ rewrite your first displayed inequality as $$ u(t) \leq \al(t)+\int_0^t \be(s) u(s) \,ds. $$ Then your last display will be $$u(t)\le e^{-2\la t}(-1+2\la(E|X_0|^2-|EX_0|^2)),$$ which is equivalent to the inequality in question (your first highlighted inequality).

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However, in the conditions of Gronwall's lemma, $\be$ is required to be nonnegative, while $\be<0$ in our setting.

Moreover, in our setting, not only Gronwall's lemma is not applicable, but also the claimed inequality $$E|X_t|^2\le\Big(|EX_0|^2+\frac1{2\la}\Big) (1-e^{-2\la t}) + E|X_0|^2 e^{-2\la t}\tag{1}\label{1}$$ will fail to hold in general even when the condition $$E|X_t|^2-E|X_0|^2\le-2\la\int_0^t E|X_s|^2 \, ds + 2t (\la|EX_0|^2+1)\tag{0}\label{0}$$ holds for all $t\ge0$.

Indeed, suppose e.g. that $\la=1$ and $$E|X_t|^2=\frac{1}{2} e^{-2 t} \left(e^t (t-2)+3 e^{2 t}+1\right)\tag{3}\label{3}$$ and $|EX_0|^2=E|X_0|^2=1$; note that the right-hand side of \eqref{3} is $\ge1$ for $t\ge0$. Then the difference between the right- and left-hand sides of \eqref{0} is $t+e^{-t} t/2\ge0$ for $t\ge0$, whereas the difference between the right- and left-hand sides of \eqref{1} is $e^{-2 t} \left(-e^t (t-2)-2\right)/2<0$ for (say) $t\ge2$. $\quad\Box$