Recreational math: If $f(f(x))=e^x$, bound the integral $\int_0^1 f(x)dx$

Question

I've been studying functions $f:\mathbb R\to\mathbb R$ that satisfy $f(f(x))=e^x$ (or, half-iterates of the exponential function). I know that there's only one such analytic function, but it's really hard to study since it is almost certainly non-elementary and I only know how to find finitely many terms of its Maclaurin Series.

Instead, I'm studying all continuous and increasing functions $f$ satisfying $f(f(x))=e^x$, and I've alighted on the following problem (which I came up with out of curiosity). I propose this question to all interested residents of MSE:

Given that $f$ is continuous and increasing and $f(f(x))=e^x$, find some bounds for the integral $$\int_0^1 f(x)dx$$

I've managed to come up with some pretty sweet bounds (in fact, they are the best possible bounds), and I'll post them after this question gets some answers.

I'll accept whatever answer has the tightest bounds, with a proof.

NOTE: Most people probably wouldn't think of this as recreational "fun" math, but hey, I did it for fun, and I'm proposing it as a problem to be done just for fun. So please try to enjoy it, and please don't try to close it.

Answer 1

Suppose $f(x)$ is an increasing, continuous function satisfying $f(f(x))=e^x$. Then there must be some $\alpha\in(0,1)$ such that $f(\alpha)=1$. We can compute $$ \begin{align*} \int_\alpha^1 f(x)\,dx&=\int_\alpha^1 e^{f^{-1}(x)}\,dx\\ &=\int_0^\alpha e^u\,df(u)\\ &=e^\alpha f(\alpha)-e^0f(0)-\int_0^\alpha f(u)e^u\,du\\ &=e^\alpha-\alpha-\int_0^\alpha f(x)e^x\,dx. \end{align*} $$ So, we have $$ \int_0^1 f(x)\,dx=\left(\int_0^\alpha+\int_\alpha^1\right) f(x)\,dx=e^\alpha-\alpha-\int_0^\alpha(e^x-1)f(x)\,dx. $$ For $0< x< \alpha$, we have $\alpha< f(x)< 1$, and thus $$ 1=e^\alpha-\alpha-\int_0^\alpha e^x\,dx<\int_0^1 f(x)\,dx<e^\alpha-\alpha-\int_0^\alpha \alpha e^x\,dx=(1-\alpha)e^\alpha+\alpha^2. $$ For $0< \alpha< 1$, the quantity $(1-\alpha)e^\alpha+\alpha^2$ takes a maximum value of $2+\log^2(2)-\log(4)$ at $\alpha=\log(2)$, so we have bounds $$ 1<\int_0^1 f(x)\,dx < 2+\log^2(2)-\log(4)\approx 1.0942. $$

These bounds are sharp: for $0\leq x\leq \log(2)$, we can take $f(x)$ to be an arbitrary continuous, strictly increasing function satisfying $f(0)=\log(2)$ and $f(\log(2))=1$, then extend $f$ to a continuous increasing function on the entire real line satisfying $f(f(x))=e^x$ by iteratively using the relation $f(x)=e^{f^{-1}(x)}$. To make $\int_0^1 f(x)\,dx$ arbitrarily close to the lower bound, we can choose $f(x)$ to be close to $1$ for an arbitrarily large portion of the interval $[0,\log(2)]$, and to make the integral arbitrarily close to the upper bound, we can choose $f(x)$ to be close to $\log(2)$ for an arbitrarily large portion of the interval $[0,\log(2)]$.

Answer 2

$\int_0^1f(x)dx=1.02525$, up to to five significant digits.

I tried to deal with this problem numerically by heavily using number crunching capabilities of Mathematica. I have simply supposed that:

$$f(x)\approx\sum\limits_{k=0}^nc_kx^k$$

Basically, you can pick $n$, calculate $f(f(x))$ and make a set of $n$ equations by matching the first $n$ coefficients of $x^k$ with the coefficients from Maclauirn's expansion of $e^x$. It's a highly non-lienar problem and I focused my efforts to find only one solution numerically. The trick is to use values $c_k$ obtained for $n-1$ as a starting point to calculate $c_k$ in the next iteration.

For $n=5$ I got something like:

$f(x)=0.503212 +0.884755 x+0.173988 x^2-0.0557584 x^3+0.17115 x^4+0.249233 x^5-0.159222 x^6-0.317672 x^7+0.172933 x^8+0.0335486 x^9$

...which is pretty bad:

Blue line represents $f(x)$, orange line represents $f(f(x))$ and green line represents $e^x$

But for $n=20$ I got something "almost" perfect:

$f(x)=-0.00110484 x^{19}+0.00423925 x^{18}-0.00285303 x^{17}-0.00653605 x^{16}+0.00430471 x^{15}+0.0114125 x^{14}-0.00188582 x^{13}-0.0167065 x^{12}-0.00616185 x^{11}+0.0170445 x^{10}+0.0165896 x^9-0.00860714 x^8-0.0207938 x^7-0.00356292 x^6+0.0140354 x^5+0.00731033 x^4+0.0207005 x^3+0.243846 x^2+0.876672 x+0.498799$

...with $\int_0^1f(x)dx=1.02526$

For $n=25$ I did not see much improvement.

I stopped at $n=30$. This is my final result:

$f(x)=3.19108553894\cdot 10^{-6} x^{29}-0.0000320371 x^{28}+0.000102301 x^{27}-0.0000852623 x^{26}-0.000155577 x^{25}+0.000202985 x^{24}+0.0003114 x^{23}-0.000304959 x^{22}-0.000659598 x^{21}+0.00024257 x^{20}+0.00120893 x^{19}+0.000250339 x^{18}-0.00172616 x^{17}-0.00138061 x^{16}+0.00165885 x^{15}+0.00293783 x^{14}-0.000399194 x^{13}-0.00403302 x^{12}-0.00203198 x^{11}+0.00346346 x^{10}+0.00438265 x^9-0.00094675 x^8-0.0047584 x^7-0.00170557 x^6+0.00287885 x^5+0.00131622 x^4+0.0240847 x^3+0.246667 x^2+0.876334 x+0.498618$

...with $\int_0^1f(x)dx=1.02525$

This value did not change even for $n=40$.

Answer 3

IMPORTANT EDIT:

I find that $f(x)$ can create some paradoxical result.

Assume $0<a<1$(which is the region of interest) and $f^{-1}(a)=b$. Then $0<b<a$.

Then, $$a=f(b)\implies \ln a=\ln f(b)=f^{-1}(b)$$

Clearly, $L.H.S.\le0$. However, we have proved that $f^{-1}(b)=R.H.S.>0$.

Can anyone resolve it?

END EDIT

Not a tight bound, but too long for a comment.

Assume the $f(x)$ is:

1. Continuous
2. A real function on the interval $(-\infty,\infty)$
   3. Strictly increasing

Since $f(x)$ is strictly increasing, it has a well-defined inverse $f^{-1}(x)$.

Let’s redefine the function as $$f(x)=e^{f^{-1}(x)}$$

Firstly, since $e^x$ is positive for all real $x$, hence $f(x)$ is strictly positive.

Secondly, we are interested in the fixed point of $f(x)$. For fixed points $(t,f(t))$, $$t=f(t)= f^{-1}(t)$$

Thus, $$t=e^t$$ which has no real solution.

Therefore, we can see $y=f(x)$ has no intersection with $y=x$.

With these two observations, one can deduce $f(x)>x$ for all $x$.

$$f(x)>x\implies \int^1_0f(x)dx>\int^1_0xdx=\frac12$$

EDIT

From $$f(x)>x$$ we have $$f(f(x))>f(x)$$ $$e^x>f(x)$$ $$e-1>\int^1_0f(x)dx$$

EDIT 2:

I think that $f(0)$ cannot be defined(but maybe okay in the sense of limit).

Assume $f^{-1}(0)=C$.

It is easy to prove $f^{-1}(x)$ is strictly increasing.

Thus, from $f(x)>x$, we get $x>f^{-1}(x)$. So $C<0$.

We also observe that, due to $f(x)>0$, $f^{-1}(x)$ is undefined for $x<0$. Since $f(x)$ is defined wherever its inverse is defined, so $f(x)$ is undefined in $(0,-\infty)$.

$$f^{-1}(0)=C\implies f(C)=0$$

But $C<0$, $f(C)$ should be undefined! Also, I have proven that $f(C)>0$(above). That leads to a contradiction.

So a value of $C$ does not exist, and hence $f^{-1}(0)$ and $f(0)$ is undefined.

From the above observations, we can also see: $$0<f^{-1}(x)<x$$.