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Let $f$ be a continuous function that is differentiable on $[a,b]$. It is known to us $f^\prime$ can be discontinuous, and $f^\prime$ can be unbounded, for instance, consider the function $f(x)=x^{3/2}\sin(1/x)$. Now my question is, how "unbound" can be say about $f^\prime$? To be more general, are there any such $f$ has a derivative that is unbounded on any subinterval $I$ of $[a,b]$?

I tried Baire's category theorem, but the best result I can obtain here is that $f^\prime$ is bounded on some set that is dense in an open interval.

MathLearner
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  • Check this: https://math.stackexchange.com/q/112067/42969 – Martin R Sep 14 '24 at 05:04
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    I see most of the discussions talking about how "discontinuous" can $f^\prime$ be, but I am interested in the unboundness. – MathLearner Sep 14 '24 at 06:56
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    See also https://math.stackexchange.com/q/3837210/42969 and the comments below the question: The continuity set of a derivative must be dense ... and continuity at a point implies there is an interval where the function is bounded. – Martin R Sep 14 '24 at 10:07

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