MHB Continuous Example: Weierstrass Function

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The discussion centers on identifying an example of an absolutely continuous function from [0,1] to ℝ that is not differentiable at infinitely many points. The Weierstrass function is initially proposed but questioned regarding its qualification as absolutely continuous. An alternative suggestion involves a piecewise continuous function connecting specific points, which could meet the criteria. The conversation highlights the importance of differentiability and continuity in the context of absolute continuity. Ultimately, the search for a suitable example continues, emphasizing the nuances of function properties in mathematical analysis.
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An example of an absolutely continuous f: [0,1] -> ℝ with infinitely many points at which f is not differentiable?

Now what I had in mind was weierstrass function which says that f(x) = Sum (n=0 to infinity) of 1/2^n cos(3^n x) and is continuous everywhere but the derivative exists nowhere...
But I am not sure if that example really fits the criteria of the question or if there might be another more suitable example of that...

Thank for the help.
 
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Why don't you just use a sine-like function, where you alternate between straight lines of slope $+1$ and straight lines of slope $-1$? You're not required to have the function differentiable nowhere, it seems. You could make this function non-differentiable at every integer. Something like this.

[EDIT]: Oops, I didn't see that the domain has to be $[0,1]$. You could probably take my function and squash it down in a limit, similar to the Weierstrass function.
 
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Juliayaho said:
An example of an absolutely continuous f: [0,1] -> ℝ with infinitely many points at which f is not differentiable?

Now what I had in mind was weierstrass function which says that f(x) = Sum (n=0 to infinity) of 1/2^n cos(3^n x) and is continuous everywhere but the derivative exists nowhere...
But I am not sure if that example really fits the criteria of the question or if there might be another more suitable example of that...

Thank for the help.

I'm not sure that the Weierstrass function qualifies as absolutely continuous.

Here's an example that should:

let $$f(x)$$ be the piecewise continuous function connecting the points $$\left( \frac{1}{2k+1},0 \right)$$ and $$\left( \frac{1}{2k},\frac{1}{2k(k+1)} \right)$$ for all integers $$k≥1$$.
 
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We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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