Prove or disprove that the function $f(x)=x^{x^{x^{x}}}$ is convex on $(0,1)$

Question

Let $0<x<1$ and $f(x)=x^{x^{x^{x}}}$ then we have :

Claim :

$$f''(x)\geq 0$$

My attempt as a sketch of partial proof :

We introduce the function ($0<a<1$):

$$g(x)=x^{x^{a^{a}}}$$

Second claim : $$g''(x)\geq 0$$

We have :

$g''(x)=x^{x^{a^{a}}+a^a-2}(a^{\left(2a\right)}\ln(x)+x^{a^{a}}+2a^{a}x^{a^{a}}\ln(x)-a^{a}\ln(x)+2a^{a}+a^{\left(2a\right)}x^{a^{a}}\ln^{2}(x)-1)$

We are interested by the inequality :

$$(a^{\left(2a\right)}\ln(x)+x^{a^{a}}+2a^{a}x^{a^{a}}\ln(x)-a^{a}\ln(x)+2a^{a}+a^{\left(2a\right)}x^{a^{a}}\ln^{2}(x)-1)\geq 0$$

I'm stuck here .

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As noticed by Hans Engler we introduce the function :

$$r(x)=x^{a^a}\ln(x)$$ We have :

$$r''(x)=x^{a^a - 2} ((a^a - 1) a^a \ln(x) + 2 a^a - 1)$$

The conclusion is straightforward the function $\ln(g(x))$ is convex so it implies that $g(x)$ is also convex on $(0,1)$.

Now starting with the second claim and using the Jensen's inequality we have $x,y,a\in(0,1)$:

$$x^{x^{a^{a}}}+y^{y^{a^{a}}}\geq 2\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{a^{a}}}$$

We substitute $a=\frac{x+y}{2}$ we obtain :

$$x^{x^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}+y^{y^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}\geq 2\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}$$

Now the idea is to compare the two quantities :

$$x^{x^{x^{x}}}+y^{y^{y^{y}}}\geq x^{x^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}+y^{y^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}$$

We split in two the problem as :

$$x^{x^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}\leq x^{x^{x^{x}}}$$

And :

$$y^{y^{y^{y}}}\geq y^{y^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}$$

Unfortunetaly it's not sufficient to show the convexity because intervals are disjoint .

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A related result :

It seems that the function :

$r(x)=x^x\ln(x)=v(x)u(x)$ is increasing on $I=(0.1,e^{-1})$ where $v(x)=x^x$ . For that differentiate twice and with a general form we have : $$v''(x)u(x)\leq 0$$ $$v'(x)u'(x)\leq 0$$ $$v(x)u''(x)\leq 0$$

So the derivative is decreasing on this interval $I$ and $r'(e^{-1})>0$

We deduce that $R(x)=e^{r(x)}$ is increasing . Furthermore on $I$ the function $R(x)$ is concave and I have not a proof of it yet .

We deduce that the function $R(x)^{R(x)}$ is convex on $I$ . To show it differentiate twice and use a general form like : $(n(m(x)))''=R(x)^{R(x)}$ and we have on $I$ :

$$n''(m(x))(m'(x))^2\geq 0$$

And :

$$m''(x)n'(m(x))\geq 0$$

Because $x^x$ on $x\in I$ is convex decreasing .

Conlusion :

$$x^{x^{\left(x^{x}+x\right)}}$$ is convex on $I$

The same reasoning works with $x\ln(x)$ wich is convex decreasing on $I$ .

Have a look to the second derivative divided by $x^x$

In the last link all is positive on $J=(0.25,e^{-1})$ taking the function $g(x)=\ln\left(R(x)^{R(x)}\right)$

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Question :

How to show the first claim ?Is there a trick here ?

Ps:feel free to use my ideas .

Answer 1

Proof:

First, we have that

(1) Midpoint convex implies rational convex

(2) And Rational convex plus continuous implies convex.

See Midpoint-Convex and Continuous Implies Convex for details on this.

So it suffices to prove the function is mid-point convex. Let $x\in (0,1)$ and let $y\in (x,1)$ then we want to show that

$$f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2}$$

That is, we want to show that

$$\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}} \le \frac{x^{x^{x^{x}}}+y^{y^{y^{y}}}}{2}$$

but this follows from the argument in your original post.

Answer 2

Mathematica 12.3 does it in moment by

NMinimize[{D[x^x^x^x, {x, 2}], x > 0 && x < 1}, x]

$\{0.839082,\{x\to 0.669764\}\}$

Since the minimum value of the second derivative on $(0,1)$ is positive, the function under consideration is convex on $(0,1)$.

Addition. The @RiverLi user states doubts concerning the NMinimize result. Here are additional arguments. First, as

D[x^x^x^x, {x, 2}] // Simplify

$ x^{x^x+x^{x^x}-2} \left(x^{2 x} \log (x) \left(x \log ^2(x)+x \log (x)+1\right)^2+x^{x+1} \log ^2(x)+x^{x+2} \log ^2(x) (\log (x)+1)^2+3 x^{x+1} \log (x) (\log (x)+1)+2 x^x+x^x \log (x) (x+x \log (x)-1)+x^{x^x} \left(x^x \log (x) \left(x \log ^2(x)+x \log (x)+1\right)+1\right)^2-1\right)$ and

D[x^x^x^x, {x, 3}] // Simplify

$x^{x^x+x^{x^x}} \left(x^{3 x} \log (x) \left(\frac{1}{x}+\log ^2(x)+\log (x)\right)^3+x^{2 x-3} \left(x \log ^2(x)+x \log (x)+1\right)^2+x^{x-2} \left(\frac{1}{x}+\log ^2(x)+\log (x)\right) \left(x^{x+1} \log (x) (\log (x)+1)+x^x-1\right)+x^{2 x^x} \left(x^x \log (x) \left(\frac{1}{x}+\log ^2(x)+\log (x)\right)+\frac{1}{x}\right)^3+3 x^{x^x-3} \left(x^{x+1} \log ^3(x)+x^{x+1} \log ^2(x)+x^x \log (x)+1\right) \left(x^{2 x+2} \log ^5(x)+2 x^x+\left(x^x+4 x-1\right) x^x \log (x)+\left(2 x^x+1\right) x^{x+2} \log ^4(x)+\left(x^{x+1}+2 x^x+2 x\right) x^{x+1} \log ^3(x)+\left(2 x^x+x+5\right) x^{x+1} \log ^2(x)-1\right)+2 x^{x-3} \left(x^2 \log ^3(x)+2 x^2 \log ^2(x)+2 x+x (x+3) \log (x)-1\right)+3 x^{2 x-3} \log (x) \left(x \log ^2(x)+x \log (x)+1\right) \left(x^2 \log ^3(x)+2 x^2 \log ^2(x)+2 x+x (x+3) \log (x)-1\right)+\frac{\left(x^{x+1} \log (x) (\log (x)+1)+x^x-1\right)^2}{x^3}+\frac{x^{x+1} \log (x)+2 x^{x+1} (\log (x)+1)+x^{x+2} \log (x) (\log (x)+1)^2-x^x+1}{x^3}+x^{x-3} \log (x) \left(x^3 \log ^4(x)+3 x^3 \log ^3(x)+3 x^2+3 x^2 (x+2) \log ^2(x)+x \left(x^2+9 x-4\right) \log (x)+2\right)\right)$ show, the second derivative is continuously differentiable on $(0,1]$. Therefore, we can draw a conclusion that the second derivative is continuously differentiable on $[0.01,1]$.

Second,

Limit[D[x^x^x^x, {x, 2}], x -> 0, Direction -> "FromAbove"]

$\infty$

and

D[x^x^x^x, {x, 2}] /. x -> 0.01

$77.923$

and

D[x^x^x^x, {x, 2}] /. x -> 1

$2$

Third, the command of Maple (here Maple is stronger than Mathematica)

DirectSearch:-SolveEquations(diff(x^(x^(x^x)), x $ 3) = 0, {0 <= x, x <= 1}, AllSolutions);

$$\left[\begin{array}{cccc} 2.58795803978585\times10^{-24} & \left[\begin{array}{c} - 1.60871316268185183\times10^{-12} \end{array}\right] & \left[x= 0.669764056702161\right] & 23 \end{array}\right] $$ shows there is only one critical point of the second derivative on $[0.01,1]$. Combining the above with the value of the second derivative at $x=0.669764056702161$, i.e. with $0.83908$, and with the result of

NMaximize[{D[x^x^x^x, {x,3}] // Simplify, x >= 0 && x <= 0.01}, x]

$\{-4779.93,\{x\to 0.01\}\},$

we conclude that the second derivative takes its global minimum on $(0,1]$ at $x=0.669764056702161$ .

Answer 3

Promising method :

As in another topic we can split in two the problem as follow :

We take the log and stop the derivative at the first order and on $\left(\frac{11}{40},1\right)$ the derivative seems to be the product of two positives increasing function wich I call $a(x)$ and $b(x)$ we have :

$$b(x)=\left[x^{\frac{615-1000}{1000}}(x^{x}(x\ln^3(x)+x\ln^2(x)+\ln(x))+1)\right]$$

And :

$$a(x)=x^{\left(x^{x}-\frac{615}{1000}\right)}$$

And :

$$\frac{d}{dx}\ln\left(x^{x^{x^{x}}}\right)=a\left(x\right)\cdot b\left(x\right)$$

So we can take the log again and again in the general case.Warning to the prohibition of calculus.

Edit :

A good substitution for $b(x)$ is $x=e^y$.Have a look at the first derivative .

It seems we have on $y\in(\ln(0.275),0)$ :

$\frac{d}{dy}b(e^y)\geq g(y)$

Where :

$g(y)=\frac{\left(e^{-\frac{385y}{1000}}\left(1000e^{3y}+e^{3y}(1000(y+1)^{2}+615)y^{3}+e^{3y}(1000(y+1)^{2}+3615)y^{2}+e^{3y}(1000\left(y+1\right)(y+3)-385)y-385e^{3y}\right)\right)}{1000}$

Or $g(y)$ equal to :

$$\frac{\left(e^{-\frac{385y}{1000}}e^{3y}\left(1000y^{5}+3000y^{4}+4615y^{3}+8615y^{2}+2615y+615\right)\right)}{1000}$$

Edit : Some ideas to show that $a(x)$ is increasing on $(0,1)$

The third derivative of $a(x)$ seems to be positive so the derivative of $a(x)$ admits a minimum .So the derivative seems to be convex as the sum of two convex function we have :

$$\left(\frac{\left(x^{x}-0.615\right)}{x}\right)''>0$$

And

$$\left(x^{x}\left(\ln^{2}\left(x\right)+\ln\left(x\right)\right)\right)''>0$$

An easier way is to substitute $x=e^y$ and then differentiate . In the derivative the embarrassing part is :

$$j(y)=e^{\left(\left(e^{y}+1\right)\cdot y\right)}\left(y\left(y+1\right)\right)+e^{y\cdot e^{y}}-0.615$$

It seems we have on $y\in(-1,0)$ :

$$j(y)\geq e^{\left(\left(y+2\right)\cdot y\right)}\left(y\left(y+1\right)\right)+e^{\left(ye^{y}\right)}-0.615\geq 0$$

Where we use the very famous inequality and $x$ a real number:

$$e^x\geq x+1$$

Again it seems that the functions :

$t(y)=e^{\left(\left(y+2\right)\cdot y\right)}\left(y\left(y+1\right)\right)$

And

$k(y)=e^{\left(ye^{y}\right)}$

Are convex for $y\in(-1,0)$ .So we can use the tangent line .