# Is there a differentiable function that approaches infinity but has a bounded derivative?

I am looking for a function $\,f : \mathbb{R} \rightarrow \mathbb{R}$ that is differentiable except at one point $x$ at which it approaches infinity. Furthermore the derivative of $\,f\,$ should be bounded in a neighborhood around $x$ and not approach infinity as one approaches $x$. Is there such a function, or is this requirement contradictory?

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It's possible if $x \in \{\pm\infty\}$, but not if $x\in\mathbb{R}$. – Michael Hardy Aug 23 '11 at 0:21

It is not possible. Fix a real number $y <x$ in the neighbournood of $x$ where $f^{\prime}$ is bounded. Since $\lim_{t \to x} f(t) = \infty,$ we may choose a sequence of $(y_n)$ of points with $y < y_n < x$ and $\lim_{n \to \infty} f(y_n) = \infty.$ Then for each $n,$ there is (by the mean value theorem), a point $u_n \in (y,y_n)$ with $f^{\prime}(u_n) = \frac{f(y_n) - f(y)}{y_n - y} > \frac{f(y_n) - f(y)}{x-y}.$ Hence we have $f^{\prime}(u_n) \to \infty$ as $n \to \infty,$ a contradiction, since each $u_n$ is in the chosen neighbourhood for each $n.$

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No, it is not possible. WLOG, assume that the point at which the function is 'infinite' is at 1. Then within a neighborhood of 1, the function takes on arbitrarily large values. Now, suppose that at 0, the value of the function is $c$. And suppose that the bound on the derivative within 2 of 1 is $K$.

Then the largest possible value for the function at 1 is $c + (1-0)K$ - but this is contradictory, as it's supposed to be infinite.

So it is not possible for such a function to exist.

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You're assuming that the derivative is integrable -- is that guaranteed to be the case? The Wikipedia article on the fundamental theorem of calculus says that the integral can be calculated from the antiderivative if the derivative is integrable -- that seems to imply that it might not be? (P.S.: The article also contains a proof that uses the assumption of integrability.) P.P.S.: Here's a counterexample: en.wikipedia.org/wiki/Volterra's_function – joriki Aug 22 '11 at 22:27
@joriki: I don't necessarily see any integration here; you can interpret the argument as using the mean value theorem for derivatives (rather than integrals): $|f(x)-f(y)| = |f'(\xi)| |x-y| \le K |x-y|$. (Of course, it's hard to tell what mixedmath really thought since it is not stated explicitly.) – Hans Lundmark Aug 23 '11 at 8:42
@Hans: I think for that to work, $f$ would have to be continuous at $x$, which it isn't; so I don't think you can avoid making the sort of limiting argument that Geoff and Shawn made. – joriki Aug 23 '11 at 8:59
@joriki: I was thinking along the mvt for derivatives, and I was sort of using the idea of a one-way continuity. As we know the function must take on an arbitrarily large values on both sides of the point of discontinuity. But the function is continuous between 0 and some $1 - \epsilon$, right? My idea was that any bound on the derivative isn't enough. But I had not considered this explicitly when I answered, and your comment made me think a lot. What do you think? – mixedmath Aug 23 '11 at 9:13
@mixedmath: Yes, spelled out like that it makes sense -- you don't really need a sequence of points going to $x$ then; you can just pick one at $x-\epsilon$ where the function value is too large to satisfy the mean value theorem, which must exist since the function values are unbounded. – joriki Aug 23 '11 at 9:20

Suppose the point where the function is not differentiable is $0$. If $\lim_{x \rightarrow 0}f(x)=\infty$, we can find an increasing sequence $x_m, x_{m+1}, \ldots$ so $\lim_{n\rightarrow \infty}x_n = 0$ and $f(x_n)=n$. Now the mean value theorem says for all $n>m$ there is a $c_n$ so $$\frac{1}{x_{n+1}-x_n}=f'(c_n).$$ This contradicts the fact that the derivative is bounded since $\lim_{n \rightarrow 0}(x_{n+1}-x_n) =0.$

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I'm not going to give a completely rigorous proof, but I'm appealing to the idea of the derivative of $f$ at $x_0$ as the slope of the line tangent to the graph of $f$ at the point $(x_0,f(x_0))$.

Up to translating right-left and up-down, we may assume that $f(0)=0$. Let $c$ such that $|f^\prime(x)|<c$ for all $x>0$ in the connected component of the domain of $f$ containing $0$. Then the graph of $f$ must be contained in the region bounded by the lines $y=\pm cx$ and as such cannot go to infinity anywhere.

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