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I recently come across some literature in stochastic analysis that uses the following result:

Consider the one-dimensional SDE

$$dX_t= a(t, X_t) \, dt + b(t, X_t) \, dW_t, $$

where $a, b: \mathbb{R}^2 \rightarrow \mathbb{R}$ are Borel-measurable functions and $\{W_t\}_{t \geq 0}$ is a Wiener process. Suppose that $a,b$ also satisfy

$1. \, \, \, \,\,\,\,\,|a(t,x)- a(t,y)| + |b(t,x) - b(t,y) | \leq K |x-y|, \quad \forall x, y \in \mathbb{R},$ $\forall t \geq 0$, for some constant $K>0$, and

$2. \, \, \, \,\,\,\,\, $the functions $t \mapsto a(t,x)$ and $t \mapsto b(t,x)$ are continuous, $\forall x \in \mathbb{R}$.

Then, the SDE has a unique strong solution.

The classical result about existence of strong solution in SDEs requires Lipschitzness in the space variable and linear growth in the space variable. I don't see how the latter can be replaced by the requirement of continuity in the time variable. Does anyone know the proof of this result?

Ref: At the bottom of Page 23 of https://spiral.imperial.ac.uk/bitstream/10044/1/28918/3/McMurray-EFV-2015-PhD-Thesis.pdf

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  • $\begingroup$ I'm confused because the continuity in time doesn't have anything to do with the behavior in the spacial direction. In particular, what if the coefficients are time-independent? Then this is saying the linear growth in frequency is unnecessary in the classical result you're referring to (which I think is very false but I could be wrong...). Even if the coefficients are time-dependent I think you still need the linear growth in frequency which is not guaranteed (I don't think) with the time-continuity condition... $\endgroup$ – Kevin Yang Jun 2 '16 at 21:42
  • $\begingroup$ @KevinYang spiral.imperial.ac.uk/bitstream/10044/1/28918/3/… Please have a look at the bottom of Page 23..... I am confused too. $\endgroup$ – Richard Jun 2 '16 at 22:39
  • $\begingroup$ "Lipschitzness in the space variable and linear growth in the space variable": should one of those "space" be "time"? $\endgroup$ – Nate Eldredge Jun 3 '16 at 5:30
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1 and 2 imply linear growth (locally in $t$). Indeed, for any $T>0$ and $t\in[0,T]$ $$ |a(t,x)|\le |a(t,0)| + |a(t,x) - a(t,0)| \le \sup_{s\in[0,T]} |a(s,0)| + K|x|\\\le \big(K\vee ||a(\cdot,0)||_{\infty;[0,T]}\big)(1+|x|). $$ As you see, it is in fact enough that $a(\cdot,0)$ and $b(\cdot,0)$ are locally bounded; with little extra effort one can prove the strong existence if they are only locally $L^1$ and $L^2$, respectively. I'll try to find a reference for the last if you need one.

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