# Help Looking for a Function with Particular Properties

I'm sorry to bother everyone, but I've been searching for functions that satisfy several properties, and so far, I've yet to be able to think of any! Specifically, the properties needed are:

$f(0)=0$ and $f(1)=1$

And on the interval $[0,1]$:

1. $f(x)$ is differentiable (though this could be relaxed a bit, the smoother, the better)
2. $f(x) \geq x$
3. $f(x)-x$ Is maximized as close to $x=1$ as possible. (Again, only between [0,1]. So, for example, for $f(x)=\sqrt{x}$, the argmax of $f(x)-x$ would be at $x=.25$ While I can find several with this value being $\leq .5$, I'm having a lot of trouble finding nice examples where this falls above $.25$!)

Where $f(x)$ really only needs to be defined over the unit interval.

I'm sorry if these properties aren't clear- just let me know and I'll try to explain better/be more specific if desired!

Thank you so much for your responses!!!

EDIT 1: Wow, you guys are astoundingly quick!! I'm so, so sorry, but I realized I (incredibly stupidly) left out two crucial properties:

1. $f(x)$ is weakly increasing on $[0,1]$
2. $0 \leq f(x) \leq 1$ (Though this can be done artificially by rescaling, so I don't think this should be an issue)

Again, I'm so sorry for forgetting to enter them into the original problem (I was trying to figure out the numbered list format, deleted the old copy, and forgot to add it back!!)

Consider the polynomial (in particular, infinitely differentiable): $$f(x)=-2.05026 x^4 + 3.74339 x^3 - 2.11177 x^2 + 1.418651 x$$

This satisfies $1\ge f(x)\ge x$ (on $[0,1]$) and has $f(x)-x$ maximized at $x\approx 0.81$.

$f(x)-x$ was found using alpha by interpolating the five points $$(0,0),(1,0),(0.9,0.05),(0.3,0.02),(0.7,0.05)$$

The first and second are problem conditions. $(0.9,0.05)$ was chosen to have a peak near $0.9$, and to have a gentle slope (between $0$ and $-1$) at the right endpoint. The other two were tweaked to keep $1\ge f(x)-x\ge 0$ for the desired interval, and to keep the rightmost peak the maximal peak.

(revised to meet updated conditions)

• I'm so, so sorry, but I forgot to enter the final condition- that $f(x)$ is weakly increasing on $[0,1]$. I'm so sorry- I got ahead of myself with the numerical list format and forgot to add it back in. Thank you so much for your time and response!! – AndrewC Apr 16 '18 at 16:55
• Solution updated to meet additional conditions. – vadim123 Apr 16 '18 at 22:47

Any concave function can be used if you scale it properly. The following works:

$f(x)=x$

$f(x)=1-(x-1)^2 = 2x-x^2$

$f(x)=sin(\pi x)$

Let $g$ be a function concave increasing on $[a,b]$ you can transform it to get a f that satisfies the properties

$\forall x \in [a,b], \ f(x)=\dfrac{g(\frac{x-a}{b-a}) -g(a)}{g(b)-g(a)}$