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It is well known that the Brownian motion is an example of functions that is continuous but nowhere differentiable. In addition, its distributional derivative can be interpreted in the way mentioned in this page.

So, how is the other functions? Can one consider the distributional derivative for functions like the Weierstrass function, Takagi function and so on? Also, am I correct in understanding that the distributional derivative of the brownian motion is a measure which is discontinuous?

Please tell me if you know.

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One can consider the distributional derivative of any locally integrable function, which includes all continuous functions. This doesn't mean there's much to say about them other than "they exist".

am I correct in understanding that the distributional derivative of the brownian motion is a measure

No, it is not a measure. The distributional derivative of a function is a signed measure if and only if the function has bounded variation on every finite subinterval. Indeed, the indefinite integral of a signed measure is the difference of two increasing functions, obtained by integrating the positive and negative components of the measure. Hence, it has bounded variation.

Since a function of bounded variation is differentiable almost everywhere, it follows that the distributional derivative of a nowhere differentiable function is not a signed measure. It is just that, a distribution. We have its definition, and there isn't much else to say.