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Is it true that a differentiable (and hence continuous) increasing bounded function $f:\mathbb{R} \to \mathbb{R}$ has derivative $f'$ that must go to zero as $x \to \infty$. If it is, could someone supply a proof?

This is just something I wanted to prove another result with, I discovered the original thing I wanted to prove was false, but I am still interested in this.

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No, this is not true. You can construct such a function like follows: Take any increasing continuously differentiable function $g$ with the following properties:

(i) $g(x)=0$ for $x\leq 0$ and $g(x)=1$ for $x\geq1$

(ii) $g'(\frac{1}{2})=1$.

You can use some scaled cosine-function as $g$ and cut the ends off. Then define $f$ by $f(x):=\frac{g(2^n(x-n))}{2^n}+\sum_{k=0}^{n-1}\frac{1}{2^k}$ for $x\in[n,n+1],n\in\mathbb N_0$ and $f(x):=0$ for $x<0$.

Since $g$ is increasing, $f$ is increasing, if it is continuous. But since $f(n)=\frac{1}{2^n}g(0)+\sum_{k=0}^{n-1}\frac{1}{2^k}=\sum_{k=0}^{n-1}\frac{1}{2^k}$ and $f(n+1)=\frac{1}{2^n}+\sum_{k=0}^{n-1}\frac{1}{2^k}=\sum_{k=0}^n\frac{1}{2^k}$ this is the case.

Since $g$ is differentiable everywhere, $f$ is differentiable in $\mathbb R\backslash\mathbb N$. Since $g$ is constant outside of $[0,1]$ we must have $g'(x)=0$ for $x\leq0$ and $x\geq1$ and this implies $f'(n)=0$ for $n\in\mathbb N$. In particular, $f$ is (continuously) differentiable even in $\mathbb N$.

Since $g\leq1$ we have $f\leq\sum_{k=0}^\infty\frac{1}{2^k}=2$, so $f$ is bounded. Since $g'(1/2)=1$ we have $f'(n+\frac{1}{2^{n+1}})=1$. In particular, $f'$ cannot go to $0$ as $x\to\infty$.

Example: Take $g(x)=-\frac{\cos(\pi x)}{2} + \frac{1}{2}$ for $x\in[0,1]$. Then $f$ looks like enter image description here

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  • $\begingroup$ Seems overly complicated to me. $\endgroup$ – zhw. Mar 13 '17 at 22:38
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Above each $n\in \mathbb N,$ put an isosceles triangle of height $1$ and base $1/2^n.$ Define $g(x)$ to equal to these triangles over those bases, with $g=0$ everywhere else. Then $g$ is continuous on $\mathbb R.$ For $x\in \mathbb R,$ define

$$f(x) = \int_0^x g(t).$$

Then $f$ is increasing and continuously differentiable on $\mathbb R,$ and $f$ is bounded above by the sum of the areas of the described triangles, which is finite. However, $f'(x) = g(x),$ which does not $\to 0$ as $x\to \infty.$

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This isn't going to be super rigorous, but hopefully somewhat convincing...

Let $f(x) = 0$ $x \in [0,1)$. Now, for $x \ge 1$, let $$f(x) = \sum_{1 \le n \le x} \frac{1}{n^2}.$$ You can extend $f(x)$ to negative $x$ by making $f(x)$ either an odd function. Then we have that $f(x)$ is bounded and increasing. However, it is not continuous. In fact, it is discontinuous for $n \in \mathbb{Z} - \{0\}$.

For $n \in \mathbb{Z} - \{0\}$, we can find small enough $\epsilon$ such that $|f(n+\epsilon) - f(n-\epsilon)| = \frac{1}{|n|^2}$. So, for each discontinuity $n$, we replace the interval $(n-\frac{1}{2|n|^2}, n+\frac{1}{2|n|^2})$ with a suitable curve that results in a continuous, differentiable function. Note that, at these intervals, $\frac{\Delta y}{\Delta x} = 1$, so these "suitable curves" that we put in must have a slope of $1$ at some point in the interval. Hence, the limit of the derivative does not exist, even though the function is continuous, differentiable, increasing, and bounded.

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