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$\newcommand\al\alpha\newcommand\be\beta\newcommand\la\lambda$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).

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 $$E|X_t|^2=\frac{1}{2} e^{-2 t} \left(e^t (t-2)+3 e^{2 t}+1\right)\tag{3}\label{3}$$ 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) $t\ge2$. $\quad\Box$

$\newcommand\al\alpha\newcommand\be\beta\newcommand\la\lambda$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).

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) $t\ge2$. $\quad\Box$

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

For real $t\ge0$, letting $$u(t):=2\la(E|X_t|^2-E|X_0|^2)-1,$$ $$\al(t):=-1,\quad\be(t):=-2\la,$$ 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 $$u(t)\le-e^{-2\la t},$$ which is equivalent to the inequality in question (your first highlighted inequality).

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 $$E|X_t|^2=\frac{1}{2} e^{-2 t} \left(e^t (t-2)+3 e^{2 t}+1\right)\tag{3}\label{3}$$ 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) $t\ge2$. $\quad\Box$

$\newcommand\al\alpha\newcommand\be\beta\newcommand\la\lambda$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).

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 $$E|X_t|^2=\frac{1}{2} e^{-2 t} \left(e^t (t-2)+3 e^{2 t}+1\right)\tag{3}\label{3}$$ 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) $t\ge2$. $\quad\Box$

$\newcommand\al\alpha\newcommand\be\beta\newcommand\la\lambda$For real $t\ge0$, letting $$u(t):=2\la(E|X_t|^2-E|X_0|^2)-1,$$ $$\al(t):=-1,\quad\be(t):=-2\la,$$ 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 $$u(t)\le-e^{-2\la t},$$ which is equivalent to the inequality in question (your first highlighted inequality).

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

For real $t\ge0$, letting $$u(t):=2\la(E|X_t|^2-E|X_0|^2)-1,$$ $$\al(t):=-1,\quad\be(t):=-2\la,$$ 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 $$u(t)\le-e^{-2\la t},$$ which is equivalent to the inequality in question (your first highlighted inequality).

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 $$E|X_t|^2=\frac{1}{2} e^{-2 t} \left(e^t (t-2)+3 e^{2 t}+1\right)\tag{3}\label{3}$$ 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) $t\ge2$. $\quad\Box$