Stochastic Differential Equations

If $C=0$ the hypothesis is $g(t)\le a=a\,e^{0\cdot t}$ and there is nothing to prove, so take $C>0$. Let $G(t)=\int_0^t g(s)\,ds$, which is absolutely continuous with $G'(t)=g(t)\le a+CG(t)$ for almost every $t$, so $\frac{d}{dt}\big(e^{-Ct}G(t)\big)=e^{-Ct}(G'-CG)\le ae^{-Ct}$ almost everywhere. Integrating from $0$ to $t$ with $G(0)=0$ gives $e^{-Ct}G(t)\le\frac aC(1-e^{-Ct})$, that is $CG(t)\le a(e^{Ct}-1)$. Substituting back, $g(t)\le a+CG(t)\le a+a(e^{Ct}-1)= a\,e^{Ct}$.

Let $X$ and $Y$ be strong solutions with the same initial value. Their difference is

the initial values cancelling. Squaring and using $(u+v)^2\le 2u^2+2v^2$, then taking expectations, the Ito isometry turns the stochastic integral's second moment into an ordinary integral and the Cauchy-Schwarz inequality bounds the drift integral,

The Lipschitz condition Equation (2) bounds each integrand by $K^2\E[(X_s-Y_s)^2]$, so with $C=2(T+1)K^2$ the function $g(t)=\E[(X_t-Y_t)^2]$ satisfies $g(t)\le C\int_0^t g(s)\,ds$, the Gronwall hypothesis with $a=0$. Here $g$ is finite and integrable on $[0,T]$, since the strong-solution definition gives $\E[X_t^2]\le 3x_0^2+3T\int_0^t\E[b^2]\,ds+3\int_0^t\E[\sigma^2]\,ds<\infty$ by Cauchy-Schwarz and the Ito isometry, and likewise for $Y$, so $g(t)\le 2\E[X_t^2]+2\E[Y_t^2]$ is bounded. By Lemma 2, $g\equiv 0$, so $X_t=Y_t$ almost surely for each $t$, and both being continuous they are indistinguishable.

Define the Picard iterates $X^0_t=x_0$ and

Each iterate is adapted and continuous, and a finite second moment propagates along the recursion. Let $A_n=\E[\sup_{s\le T}(X^n_s)^2]$. Then $A_0=x_0^2<\infty$, and if $A_n<\infty$ then linear growth gives $\int_0^T\E[\sigma(s,X^n_s)^2]\,ds\le K^2T(1+A_n)<\infty$, so the Ito integral in Equation (5) is a genuine $L^2$ martingale and both the Ito isometry and Doob's $L^2$ inequality apply; bounding $X^{n+1}$ by $(u+v+w)^2\le 3u^2+3v^2+3w^2$, Cauchy-Schwarz on the drift, and Doob on the martingale, $A_{n+1}\le 3x_0^2+3TK^2T(1+A_n)+12K^2T(1+A_n)<\infty$. By induction every $A_n$ is finite, so each integral is square-integrable and the recursion is well defined. Write $\Delta^n_t=X^{n+1}_t-X^n_t$ and $\varphi_n(t)=\E\big[\sup_{s\le t}(\Delta ^n_s)^2\big]$, which the bound $A_n<\infty$ renders finite. For $n\ge 1$, the increment $\Delta^n$ is Equation (3) with $X^n,X^{n-1}$ in place of $X,Y$, so by $(u+v)^2\le 2u^2+2v^2$, the Cauchy-Schwarz bound on the drift, and Doob's $L^2$ inequality with the Ito isometry on the martingale part,

the factor $8$ being twice Doob's constant $4$. The Lipschitz bound makes both integrands at most $K^2\E[(\Delta^{n-1}_s)^2]\le K^2\varphi_{n-1}(s)$, so with $C=2(T+4)K^2$,

Since $\varphi_0(T)=\E[\sup_{s\le T}(X^1_s-x_0)^2]$ is a finite constant $M$ by linear growth, iterating Equation (7) gives $\varphi_n(t)\le M\,(Ct)^n/n!$, whose square roots sum to a finite value, $\sum_n\varphi_n(T)^{1 /2}\le\sum_n\sqrt{M(CT)^n/n!}<\infty$, making the iterates a Cauchy sequence in the complete space of adapted processes with finite $\E[\sup_{s\le T}(\cdot)^2]$, and converge uniformly in mean square to a limit $X$, which is adapted and has a continuous version, since a subsequence converges uniformly in $t$ almost surely.

Passing to the limit in Equation (5) identifies $X$ as a solution. The drift integral converges because $\int_0^t(b(s,X^n)-b(s,X))\,ds$ has mean square at most $T K^2\int_0^t\E[(X^n_s-X_s)^2]\,ds\to 0$ by Lipschitz continuity, and the stochastic integral converges by the Ito isometry and the same Lipschitz bound. The mean-square convergence gives $\E[\sup_{s\le T}X_s^2]<\infty$, so linear growth yields $\int_0^T\E[\sigma(t,X_t)^2]\,dt\le K^2T(1+\E[\sup X^2])<\infty$ and likewise for $b$, the integrability required by Definition 1. So $X_t=x_0+\int_0^t b(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dW_s$, a strong solution.