Is there a bounded, increasing elementary function with unbounded derivative?

Question

Is there an elementary function function $f(x)$ with the following properties:

• $f(x)$ is defined and infinitely differentiable on all of $\mathbb{R}$
• $f(x)$ is bounded
• $f^\prime(x)$ is nonnegative and unbounded

If we don't require $f(x)$ to be an elementary function, then we can take $$f(x)=\int_0^x\frac{t^2}{1+t^8\sin^2(\pi t)}\,dt$$ (the integrand is nonnegative, unbounded, but has finite area, desmos link).

This question can also be rephrased in terms of an unbounded probability mass function with elementary antiderivative.

Answer 1

It suffices to find a function with those properties on half of $\mathbb R$. Indeed, if

$$g:[0,+\infty)\to\mathbb R$$

is bounded and $g'\geq0$ is unbounded (and let's also assume $g(0)=0$ and $\lim_{x\to\infty}g(x)=1$, so $g(x)\geq0$ for all $x\geq0$), then

$$f:\mathbb R\to\mathbb R$$ $$f(x)=\tanh(x)\cdot g(x^2)$$ $$f'(x)=\text{sech}^2(x)\cdot g(x^2)+2x\tanh(x)\cdot g'(x^2)$$

has the same required properties. So let's try to find such $g$.

Since $g$ must be increasing and bijective from $[0,\infty)$ to $[0,1)$, we can write

$$g(x)=\tanh\!\big(h(x)\big)$$

for some function $h$ which is increasing and bijective from $[0,\infty)$ to $[0,\infty)$. Translating the condition on $g'$ to a condition on $h'$,

$$g'(x)=\text{sech}^2\big(h(x)\big)\cdot h'(x)$$ $$\frac{h'(x)}{e^{2h(x)}}$$

must be unbounded.

Let's make the assumption that $h(x)\approx x$, so the denominator simplifies to $e^{2x}$. We need $h$ to be smooth, like $x+\sin(x)$, but getting sharper as $x$ increases, so that it more resembles $\lfloor x\rfloor$. Thus, let

$$h(x)=x+i(x)$$

for some bounded function $i$. We must have $i(0)=0$, and $i'(x)\geq-1$, and

$$\frac{i'(x)}{e^{2x}}$$

must be unbounded.

Considering the sawtooth wave $\arctan(\tan(x))$ as our prototype, let's try the form

$$i(x)=-\arctan\left(\frac{\sin(x)\cos(x)}{\cos^2(x)+j(x)^2}\right)$$

where $j(x)>0$ but $\lim_{x\to\infty}j(x)=0$. The derivative is

$$i'(x)=-\frac{\cos(2x)\big(\cos^2(x)+j(x)^2\big)-\tfrac12\sin(2x)\big(-\sin(2x)+2j(x)j'(x)\big)}{\sin^2(x)\cos^2(x)+\big(\cos^2(x)+j(x)^2\big)^2}.$$

Evaluating this where $\cos(x)=0$ (i.e. at the vertical edges of the sawtooth wave, where $i'$ is supposed to explode),

$$i'(x_n)=\frac{1}{j(x_n)^2},\qquad x_n=\frac\pi2+n\pi,\quad n\in\mathbb N.$$

So, for the derivative of $g$ to be unbounded, it suffices to have $e^{x_n}j(x_n)$ approach $0$.

For general $x$, we must have $i'(x)\geq-1$; this inequality simplifies to

$$-\sin(2x)\,j(x)j'(x)\leq j(x)^2+j(x)^4$$

which is equivalent to

$$\sin(2x)\,\frac{d}{dx}\ln\left(\frac{\sqrt{1+j(x)^2}}{j(x)}\right)\leq1.$$

Let's call this function being differentiated $k(x)$. Thus

$$j(x)=\frac{1}{\sqrt{e^{2k(x)}-1}}=\frac{e^{-k(x)}}{\sqrt{1-e^{-2k(x)}}}.$$

Noting that $j(x)\to0$ is equivalent to $k(x)\to\infty$, the unboundedness condition translates to $k(x_n)-x_n\to\infty$.

The other condition $i'(x)\geq-1$ becomes $\sin(2x)\,k'(x)\leq1$. An obvious attempt at a solution is $k'(x)=1$, so $k(x)=x$, but that doesn't satisfy the unboundedness condition. Let's try to make $k'(x)$ oscillate, so that it's greater than $1$ on average, but less than $1$ where $\sin(2x)$ is large:

$$k'(x)=\frac54-\frac34\sin(2x)$$ $$\sin(2x)\,k'(x)=\frac{25}{48}-\frac34\left(\sin(2x)-\frac56\right)^2\leq\frac{25}{48}\leq1$$ $$k(x)=\frac54x+\frac38\cos(2x)$$ $$k(x_n)-x_n=\frac14x_n+\frac38\cos(2x_n)\to\infty.$$

So we have a solution!

---

$$k(x)=\frac54x+\frac38\cos(2x)$$ $$j(x)=\big(e^{(5/2)x+(3/4)\cos(2x)}-1\big)^{-1/2}$$ $$i(x)=-\arctan\left(\frac{\sin(x)\cos(x)}{\cos^2(x)+(e^{(5/2)x+(3/4)\cos(2x)}-1)^{-1}}\right)$$ $$h(x)=x-\arctan\left(\frac{\sin(x)\cos(x)}{\cos^2(x)+(e^{(5/2)x+(3/4)\cos(2x)}-1)^{-1}}\right)$$ $$g(x)=\tanh\left(x-\arctan\left(\frac{\sin(x)\cos(x)}{\cos^2(x)+(e^{(5/2)x+(3/4)\cos(2x)}-1)^{-1}}\right)\right)$$ $$f(x)=\tanh(x)\tanh\left(x^2-\arctan\left(\frac{\sin(x^2)\cos(x^2)}{\cos^2(x^2)+(e^{(5/2)x^2+(3/4)\cos(2x^2)}-1)^{-1}}\right)\right)$$

Answer 2

As mentioned in my comment, this is only a partial answer, as instead of $f'$ unbounded I have $f''$ unbounded. OP has also specified that they would accept finite differentiability almost everywhere.

Let $$ f'(x) = \frac{4x^{2/3}}{1+x^2} $$It's easily seen that $$ f''(x)=-\frac{8\left(2x^2-1\right)}{3x^{1/3}\left(x^2+1\right)^2}, $$which is unbounded as $x\to 0$. To see that $f(x)$ is elementary, make the substitution $z=x^{1/3}$, or $z^3=x$: $$ f(x) = \int \frac{4x^{2/3}}{1+x^2} dx $$ $$ \stackrel{z^3=x}{\Rightarrow} 12\int \frac{z^4}{1+z^6}dz $$Since the integrand is a rational function, its primitive is an elementary function, and this is still true when we back-substitute for $x$. Taking $C=0$, a closed-form is $$ f(x)=\sqrt{3}\log\left(\frac{x^{2/3}-\sqrt{3}x^{1/3}+1}{x^{2/3}+\sqrt{3}x^{1/3}+1}\right)+2\arctan(2x^{1/3}-\sqrt{3})+2\arctan(2x^{1/3}+\sqrt{3})+4\arctan(x^{1/3}) $$