It is possible for a derivative to fail to exist at isolated points, but I would like to know if a function could be constructed that is not differentiable almost everywhere and differentiable at isolated points only.
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1https://collegemathteaching.wordpress.com/2014/02/24/a-real-valued-function-that-is-differentiable-at-an-isolated-point/ – parsiad Dec 09 '16 at 04:22
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1You could squeeze a chaotic function with eg. $\sin^2 x$ – Joffan Dec 09 '16 at 04:22
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1@Joffan $\sin^2(x)$ is still differentiable on $\mathbb{R}$, so transformations of it should be also. How would you use $\sin^2(x)$to construct a function differentiable only at isolated points? – user56202 Jan 08 '21 at 14:56
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@user56202 the point is exactly as described in the accepted answer; $\sin^2 x$ could act as the multiplying "squeezer" to impose differentiability to a chaotic function at selected points. – Joffan Jan 08 '21 at 15:04
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@Joffan Ah I see, you mean $\sin^2(x) g(x)$ where $$g(x) = \begin{cases} 1 & \textrm{if $x \in \mathbb{Q}$} \ 0 & \textrm{if $x \notin \mathbb{Q}$}. \end{cases}$$ whereas the accepted answer used $x^2 g(x)$. – user56202 Jan 08 '21 at 15:33
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@user56202 I didn't specify which chaotic function, just the squeezing concept. And my comment was made about 15 minutes before the fuller exposition in the accepted answer was published. – Joffan Jan 08 '21 at 17:27
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@Joffan Yes I understand, thank you for your replies. – user56202 Jan 08 '21 at 17:55
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Consider $$f(x) = \begin{cases} x^2 & \textrm{if $x \in \mathbb{Q}$} \\ 0 & \textrm{if $x \notin \mathbb{Q}$}. \end{cases}$$ Observe that $f$ is continuous only at $x = 0$, hence $f$ is not differentiable except possibly at $x = 0$. We claim that $$\lim_{x \to 0}\frac{f(x)}{x} = 0,$$ i.e. $f'(0) = 0$. To see this, note that $\tfrac{f(x)}{x} = 0$ for all $x \not\in \mathbb{Q}$. If $x \in \mathbb{Q} \setminus \{0\}$, then $\tfrac{f(x)}{x} = x,$ hence if $\vert x \vert < \epsilon$, then $\left\vert \tfrac{f(x)}{x} \right\vert < \epsilon$.
If you want continuity, you might be interested in the Weierstrass function $$f(x) = \sum_{n=0}^{\infty}a^n\cos(b^\pi x)$$ where $0 < a < 1$, $b$ is a positive odd integer, and $ab > 1 + \tfrac{3}{2}\pi$. It can be shown that $f$ is continuous everywhere but differentiable nowhere.
Now if you want an example of a function continuous everywhere but differentiable at only a point, take $f$ to be a continuous everywhere but differentiable nowhere function (i.e. the Weierstrass function), then define $g(x) = x^2f(x)$. Since $f(x) = \tfrac{g(x)}{x^2}$ is differentiable nowhere, $g$ cannot be differentiable except possibly at $x = 0$. You can easily show that $g'(0) = 0$.
Ethan Alwaise
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Very interesting. Then I wonder: If I define a polynomial $q(x) > 0$ (I assume that positivity is needed to remove the possibility of two-sided limits not existing?) then define $g(x) = q(x)f(x)$ and the same reasoning holds at all the zeros of $g(x)$, so $f(x)$ is differentiable only at the zeros of $q(x)$? If that's the case, what if $q(x) = \sum_{i=0}^\infty (x - q_i)^2$, where $q_i$ is the $i^{th}$ rational number? This polynomial is infinite, but would the same reasoning hold? (I'm trying to see if we can make $f$ differentiable at an infinite number of isolated points.) – Michael Stachowsky Dec 09 '16 at 11:55
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As an update to what I just wrote, I realized that using $q(x)$ would not necessarily be the way to go, since its limits are infinity (everywhere?). Instead, let $q(x) = sin(x)^2$, which would have a zero derivative at an infinite number of points along $\mathbb{R}$ but would also be finite everywhere. – Michael Stachowsky Dec 09 '16 at 13:33
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1I don't get how the Rationals are isolated points... nor the irrationals... What are you using as a definition of isolated point? – An old man in the sea. Sep 05 '19 at 17:41