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Assume $f$ is continuous and differentiable on $[a,b]$, can $f$ has a point where its derivative goes to infinity within that interval? In other words, if $f$ is continuous and differentiable on a finite interval, can $f'$ has a singularity point on that interval?

cyberic
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  • "In other words" doesn't seem accurate. The second sentence would allow for an oscillatory discontinuity, but the first not. Do you just want an example of a discontinuous derivative, or must it be an infinite discontinuity? – saulspatz Mar 26 '18 at 13:45
  • @saulspatz Actually I'm curious to know an example of a discontinuous derivative now – cyberic Mar 26 '18 at 13:55
  • Consider an upper semicircle with radius $1$ centered at $(1,0)$ and a lower semicircle with radius $1$ centered at $(3,0)$. The derivative will be discontinuous at $x=2$. – ThePortakal Mar 26 '18 at 14:35
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    Please make the question more precise: what exactly do you mean by a "singularity", a point "where its derivative goes to infinity"? To say that $f$ is differentiable on $[a,b]$ just means that for all $x\in[a,b]$, $f'(x)$ exists (meaning, it is a real number). In that sense, $f'$ cannot have a singularity. However, it is possible that $f'(x)$ exists for all $x\in[a,b]$, even though $f'(x_0)\to\infty$ for some $x_0\in[a,b]$. This requires the derivative to be a discontinuous Darboux function (a discontinuous $f'$ that nevertheless satisfies the Intermediate Value Property). – symplectomorphic Mar 26 '18 at 15:17

3 Answers3

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In fact, it is possible for a function to be differentiable but to have an unbounded derivative. An answer to this question gives the example $$f(x)=\cases{ x^2\sin(1/x^2),&$x\ne0$ \cr 0,&$x=0$}$$ for $-1\le x\le 1.$ It is easy to confirm this: $$f'(x)=\cases{2x\sin \frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2},&$x\ne0$ \\0,&$x=0$}$$ where $f'(0)$ is computed from the definition of derivative.

Note that $\lim_{x\to 0-}f'(x)=+\infty, \lim_{x\to 0+}f'(x)=-\infty.$ If we define $$g(x)=\cases{-f(x),&$x\ge0$ \\f(x), &$x<0$}$$ then $lim_{x\to 0 } g(x)=+\infty.$

saulspatz
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  • OP wants to know whether $f'(x)$ exists and it is finite. The question is not about whether $f'(x)$ is continuous. Your example shows that $f'(x)$ is not continuous but for each $x\in[-1,1]$ we have $f'(x)$ is well defined and finite. – Arian Mar 26 '18 at 14:08
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    @Arian So what does his comment, "Actually I'm curious to know an example of a discontinuous derivative now," mean? Also, a continuous function on a closed interval is bounded. If the derivative goes to $\infty$ it is necessarily discontinuous. – saulspatz Mar 26 '18 at 14:15
  • Yes, your example answers his comment. This is true. However the original question is different. The question is about if $f'(x)$ exists and is finite for all $x$ under given conditions. – Arian Mar 26 '18 at 14:19
  • @Arian I disagree. In this example, $f'(x)$ exists and is finite, at all points and goes to to $\infty$ at some point. Are you objecting because the one-sided limits are different? That is trivial to fix. I've made this change. – saulspatz Mar 26 '18 at 14:27
  • No, my point here is not whether right and left limit of $f'(x)$ agree or not. Because the original question is not about continuity of $f'(x)$. It is merely about calculating $f'(x)$ using the definition of the derivative and find out whether it is finite for $x$ in the interval. Surely if the question was about continuity (or boundedness) of $f'(x)$ your answer is absolutely correct. I agree with this. – Arian Mar 26 '18 at 14:34
  • @Arian: The OP said "where its derivative goes to infinity", which is what saulspatz gives. I looked at the original version and even there "unbounded derivative" appears, which is what the OP is asking for (in a different way). The fact that $f'(x)$ is also discontinuous is mostly irrelevant. Indeed, $f'(x)$ being discontinuous would not be enough, since there exists discontinuous derivatives whose derivatives are bounded. But the example given here is not one of those --- the example given here actually has an unbounded derivative. – Dave L. Renfro Mar 26 '18 at 14:40
  • @DaveL.Renfro I think OP asks two different questions. The first one is what saulspatz answers. The second one is what I answer since a singularity point $x$ for a function $f$ would mean $f(x)=\pm\infty$ I suppose. Anyways. – Arian Mar 26 '18 at 14:52
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    @Arian: You might be right. Of course, when someone says differentiable (presumably finitely differentiable is meant) on $[a,b],$ this means that every point the derivative is defined and finite, which would exclude infinite singularities. But maybe the OP is still at the stage where these two concepts are sometimes conflated. – Dave L. Renfro Mar 26 '18 at 15:05
  • @Arian I feel that this answer contradicts your answer (I also feel from your discussion it's not) Let me be specific and more mathematical rigorous (I will do my best) My understanding so far, that $f' < $ Infinity on $[a,b]$ if $f$ is differentiable on $[a,b]$ where I got this from your answer. From this answer, I understand that $f'$ can go to infinity even $f$ is differentiable on $[a,b]$ – cyberic Mar 26 '18 at 15:32
  • @cyberic yes. they are two different things. and this is an important and interesting point about differentiability. The function can have finite derivative at each point in the interior of its domain, but yet have a discontinuous derivative. – Arian Mar 26 '18 at 15:37
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by definition if a function $f$ is continuous on a compact interval $[a,b]$ and differentiable on the open interval $(a,b)$ then in particular the following limits exists and are equal: $$\lim_{y\to x^+}\frac{f(y)-f(x)}{y-x}=\lim_{y\to x^-}\frac{f(y)-f(x)}{y-x}:=f'(x)$$ for each $x\in (a,b)$. This automatically excludes the possibility that there is such an $x\in (a,b)$ such that $f'(x)=\pm\infty$ otherwise it means the limits above don't exist.

Arian
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    This doesn't show that it can't happen that $f'(x_n)\to\infty$ for some $x_n$ –  Mar 26 '18 at 13:48
  • @acetone if $f$ is continuously differentiable (its derivative is continuous) then $f'(x_n)\to f'(x)$ whenever $x_n\to x$. In the case when $f'(x)$ is not continuous then I suspect the only type of discontinuities are jump discontinuities and situations $f'(x_n)\to\pm\infty$ do not arise as $x_n\to x$. – Arian Mar 26 '18 at 13:56
  • That is not given. Also, derivatives cannot have jump discontinuities. Only essential ones. If a derivative has a lateral limit, then it must be continuous from that side. –  Mar 26 '18 at 14:01
  • Anyways OP wants to know whether $f'(x)$ exists and it is finite which indeed it is the case. The question is not about whether $f'(x)$ is continuous. – Arian Mar 26 '18 at 14:09
  • This is incorrect. See the example I cite in my answer. – saulspatz Mar 26 '18 at 14:09
  • @saulspatz your example shows that $f'(x)$ is not continuous at $x=0$. However it clearly shows that for each $x$ in your interval $f'(x)$ exists and it is finite. No? – Arian Mar 26 '18 at 14:12
  • @Arian That is correct, but the derivative is unbounded, as the OP requests. – saulspatz Mar 26 '18 at 14:17
  • You just misunderstood the question and are trying to answer the question: "If a function is differentiable then its derivative exists." which is just the definition of the word differentiable. All those lines and formula you wrote are unnecessary for answering your misinterpretation of the question. The question is actually, the one with real content: Whether $f'$ is bounded. –  Mar 27 '18 at 01:49
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For piecewise continuity, we can certainly have that there is a singularity point. But for strict continuity then this would not be true since if we let $f$ be continuous on the finite interval $x\in [a,b]$, with $a<b$. Then we know $c_1\leq f \leq c_2$ , so we can write:

$$\frac{|f(x)-f(y)|}{x-y}\leq\frac{c_2-c_1}{x-y}$$ By the M.V.T, there exists a $\hat{c}$ in $[x,y]$ such that $f'(\hat{c})=\frac{c_2-c_1}{x-y}$ which is finite.

Btzzzz
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