Putname Competition: Aftermath B?

No such function exists. I will prove it by contradiction. The longest part of my proof is showing that f ' is bounded. Once I finish that, the proof is very short. Maybe you can think of better ways to do it.

Proof that f ' is bounded:

Suppose that there is a strictly increasing function f : R --> R such that f '[x]=f[f[x]] for all x in R (so f is differentiable). Then since f is strictly increasing, f '[x]>0 for all x, and looking at what f '[x] is equal to, it is differentiable with derivative f ' '[x]=f '[f[x]]*f '[x]>0. So f is convex.

Since f is convex, it lies above all of its tangent lines, so for each b we get that

f[x]>=f[b]+f '[b]*(x-b) for all real x. This line has positive slope, so it is eventually positive, and hence f is eventually positive. Fix B such that f[b]>0 for all b>B.

By the convexity relation above, at x=f[b] we have f[f[b]]>=f[b]+f[f[b]]*(f[b]-b), so

f[f[b]]*(1+b-f[b])>=f[b]. For all b>B, f[b]>0, so f[f[b]]*(1+b-f[b])>0, and f[f[b]]=f '[b]>0 implies that 1+b-f[b]>0. So f[b]-b<1 for all b>B. The function f[b]-b is convex (its second derivative is the same as that for f), so all of its tangent lines must have 0 or negative slope (if one had positive slope, f[b]-b would be above a line with positive slope, and hence would not be bounded above by 1 for large b). Thus, (f[b]-b)'=<0 for all b, which we can write as f '[b]=<1 for all b. Then f '[x] is in (0,1] for all x in R.

Completing the proof that a function with these properties cannot exist:

But f ' is also convex, since f ' ' '[x] = (f '[f[x]]*f '[x])' = f ' '[f[x]]*f '[x]^2+f '[f[x]]*f ' '[x]>0 for all x. So since f ' lies above its tangent lines, the slope f ' '[x] must be 0 at all x (since f ' is bounded above; the slope cannot even be negative this time, since for large negative x this would make f ' >1). This contradicts that f ' '[x]=f '[f[x]]*f '[x]>0 for all x.

Thus, no such function exists.

Since f is increasing, f' is increasing and so f is convex. There exist 2 constants A > 0 and B so that

f(x) > Ax + B, hence f'(x) = f(f(x)) > A^2 x + C and f(x) > A^2 x^2 /2 + Cx + D.

Now f'(x) / f(f(x) ) = 1, so it's integral diverges at infinity but

int_[0,inf] f'(x) / f(f(x) ) dx = int [f(0), inf] 1/ f(x) dx.

The divergence of this integral contradicts f(x) > A^2 x^2 /2 + Cx + D.