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We have a well-defined SDE: $$ {\rm d}X_t=\mu(X_t){\rm d}t+\sigma(X_t) {\rm d}B_t, $$ where the initial condition $X_0$ is a known r.v., and $B_t$ is a standard Brownian motion.

Can we say the above SDE is a "continuous"-mapping $f:\mathbb{R}\times C[0,\infty)\to C[0,\infty)$ which maps $(X_0,B_t)$ into $X_t$ ?

What conditions should we need to realize the above statement? Could you provide some literature?

Thank you very much!

Mike HWANG
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    Your mapping $f$ is called the Itô map. When a (pathwise-)unique strong solution to the SDE exists it is known that $f$ is measurable however in dimensions 2 and higher it is generally not continuous. Using the theory of rough paths, one can factor $f$ into a "lift" into new space and a continuous map from that space into $C\left([0, \infty)\right)$. A good reference for rough paths is Friz and Hairer's A Course on Rough Paths. – Shiva Feb 18 '23 at 12:08
  • @Shiva Thank you for your insightful answer ! Can I understand that Rough Path Theory use a different metric for the space $C[0,\infty)$ which is not the well-known uniform metric ? – Mike HWANG Feb 19 '23 at 13:25

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