I am trying to find a single (i.e. non-piecewise) analytic function $f(x)$ with the following features:

$f(x), f'(x)$ defined $ \forall x \in \mathbb R$
$f'(x) > 0, \forall x$
$lim_{x \to - \infty} f(x) = - \infty$
$lim_{x \to + \infty} f(x) = + \infty$
$lim_{x \to - \infty} f'(x) = m, m \in \mathbb R^+$
$lim_{x \to + \infty} f'(x) = m, m \in \mathbb R^+$
$f'(\frac 1 2) = n, n \in \mathbb R^+, n \le \frac m {10}$ # no longer applicable, see edit below #

So a monotonically increasing curve that approximates a straight line of slope $m$ almost everywhere, except for a section with much lower slope $n$ around $x = \frac 1 2$, say for $0.1 < x < 0.9$.

How would you go about putting together such a function?


I had forgotten to include a condition:

$f'(x) \ge n, \forall x$

Also, I realized the function I am looking for becomes symmetric with respect to the origin if I translate $x$ by $- \frac 1 2$, so in fact the previously mentioned condition on the derivative becomes:

$f'(0) = n, n \in \mathbb R^+, n \le \frac m {10}$


EDIT: Taking into account the OP's edits, we have to modify the previous function slightly. I'm merging it with the original answer for clarity.

Assume $m,n,a > 0, m>n$. Then, consider the function $$f(x) = \int_0^x \left(m-(m-n)e^{-\frac{u^2}{a^2}}\right) du = mx-(m-n)a\frac{\sqrt{\pi}}{2} \ \text{erf}\left(\frac{x}{a}\right)$$

Thus, we get for $f(x)$

  1. As $\text{erf}(-\infty)$ is finite, $\lim_{x\to-\infty} f(x) = -\infty$
  2. As $\text{erf}(\infty)$ is finite, $\lim_{x \to \infty} f(x) = \infty$

Also, we have $f'(x) = m-(m-n)e^{-\frac{x^2}{a^2}}$. This gives for $f'(x)$

  1. For all $x$, as $e^{-\frac{x^2}{a^2}} \le 1$, we get \begin{align} f'(x) &= m-(m-n)e^{-\frac{x^2}{a^2}} \\ &\ge m-(m-n)\\ &=n \end{align}
  2. $f'(0) = m-(m-n)e^{-\frac{0^2}{a^2}}= m-(m-n) =n \ \ \forall \ x$
  3. $\lim_{x\to-\infty} f'(x) = \lim_{x\to-\infty} \left(m-(m-n)e^{-\frac{x^2}{a^2}}\right) = m$
  4. $\lim_{x\to\infty} f'(x) = \lim_{x\to\infty} \left(m-(m-n)e^{-\frac{x^2}{a^2}}\right) = m$

Note that $m, n, a$ here are completely general. So, we can now impose any other conditions, like $n\le \frac{m}{10}$ trivially.

  • $\begingroup$ That's brilliant, thank you very much! I realize now that I omitted an important condition: $f'(x) \ge n, \forall x$. And in fact I would need $m$ to be a parameter in the equation. I might try to adapt your equation to achieve this, but if you have any suggestions, as you were so quick finding what I was looking for... $\endgroup$ Jan 22 at 18:12
  • $\begingroup$ Could you write it down in your main question? It'll both help me and others to answer your question better. $\endgroup$
    – Ishan Deo
    Jan 22 at 19:29
  • $\begingroup$ Sure, thanks. Done. $\endgroup$ Jan 22 at 19:42
  • $\begingroup$ There. I've changed it. $\endgroup$
    – Ishan Deo
    Jan 22 at 20:07

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