This seems like it should be true but I was wondering if anyone could prove it. Thanks!
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2$\begingroup$ Start with a continuous nowhere-differentiable function on $\mathbb{R}$ and multiply by $x^2$. Now it's differentiable at $x=0$. $\endgroup$– Nik WeaverCommented Nov 14, 2023 at 11:28
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The answer is negative. Take a bounded nowhere differentiable continuous map $f : \mathbb{R} \to \mathbb{R}$ and consider the map $g(x) = f(x) \cdot x^2$, which is differentiable at $x = 0$, but not anywhere else.
answered Nov 14, 2023 at 11:24
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$\begingroup$ Thanks! I have one more question, how about Lipschitz continuous function which is differentiable almost everywhere? If a Lipschitz continuous function is differentiable at a point, is it differentiable in some neighborhood around that point? $\endgroup$ Commented Nov 14, 2023 at 11:48
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2$\begingroup$ I think these questions are better suited for math.stackexchange.com $\endgroup$ Commented Nov 14, 2023 at 11:51
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$\begingroup$ @li ang Duan: If a Lipschitz continuous function is differentiable at a point, is it differentiable in some neighborhood around that point? --- A result Zahorski published in 1946 takes care of this and most any related question. The result, along with some very strong specific examples, is given in this MSE answer. See also this paper. $\endgroup$ Commented Nov 14, 2023 at 20:44