# Mean Square Differentiability

If I want to show that a stochastic process is not mean square differentiable, is it enough to show, that the process $a.s.$ does not have differentiable sample paths?

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Found the following statement:

Since mean square differentiability is generally straightforward to determine, the more difﬁcult cases involve showing sample path differentiability for processes that are mean square differentiable. Doob (1953, p. 536) shows that for a separable process, sample functions of the process are absolutely continuous, and hence the functions have derivatives almost everywhere.

So if your process is sepearable and the sample paths aren't differentiable almost everywhere, the process won't be mean square differentiable.

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