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Nick Mendler
Nick Mendler
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The derivative of any nowhere continuously differentiable bounded variation (BV) function will work.

If $f$ is BV, then it is differentiable a.e., but if its derivative were anywhere continuous, then $f$ would be continuously differentiable at that point. Thus every $g\in [f']$ must be everywhere discontinuous.

You can get plenty of examples of BV functions that are nowhere continuously differentiable by considering stochastic processes. This is also good applied motivation.

If $H$ is a bounded random variable, and $Y_t = \int_0^t H dX$ is defined by a stochastic integralSome fractal curves will have this property, then $Y_t$ is necessarily BVsee Mandelbrot and almost surely nowhere continuously differentiable. Hence the derivativeFrames "canopy of $Y_t$ is almost surely discontinuous everywherea self contacting fractal tree".

The derivative of any nowhere continuously differentiable bounded variation (BV) function will work.

If $f$ is BV, then it is differentiable a.e., but if its derivative were anywhere continuous, then $f$ would be continuously differentiable at that point. Thus every $g\in [f']$ must be everywhere discontinuous.

You can get plenty of examples of BV functions that are nowhere continuously differentiable by considering stochastic processes. This is also good applied motivation.

If $H$ is a bounded random variable, and $Y_t = \int_0^t H dX$ is defined by a stochastic integral, then $Y_t$ is necessarily BV and almost surely nowhere continuously differentiable. Hence the derivative of $Y_t$ is almost surely discontinuous everywhere.

The derivative of any nowhere continuously differentiable bounded variation (BV) function will work.

If $f$ is BV, then it is differentiable a.e., but if its derivative were anywhere continuous, then $f$ would be continuously differentiable at that point. Thus every $g\in [f']$ must be everywhere discontinuous.

Some fractal curves will have this property, see Mandelbrot and Frames "canopy of a self contacting fractal tree".

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Nick Mendler
Nick Mendler
  • 103
  • 1
  • 4

The derivative of any nowhere continuously differentiable bounded variation (BV) function will work.

If $f$ is BV, then it is differentiable a.e., but if its derivative were anywhere continuous, then $f$ would be continuously differentiable at that point. Thus every $g\in [f']$ must be everywhere discontinuous.

You can get plenty of examples of BV functions that are nowhere continuously differentiable by considering stochastic processes. This is also good applied motivation.

If $H$ is a bounded random variable, and $Y_t = \int_0^t H dX$ is defined by a stochastic integral, then $Y_t$ is necessarily BV and almost surely nowhere continuously differentiable. Hence the derivative of $Y_t$ is almost surely discontinuous everywhere.