Hi I want to show the following fact :
Problem :
Let $x\geq 1$ and $n\geq 1$ a natural number and define:
$$f(x)={}^{2n}x=\underbrace{x^{x^{⋰^{x}}}}_{2n\text { times}}$$
Then we have :
$$f''(x)\ge 0$$
My sketch of proof for $x\ge y\geq 1$ :
We suppose by induction that $k\geq 2$ a natural number :
$${}^{k-1}x+{}^{k-1}y\geq {}^{k-1}\left(\frac{x+y}{2}\right)*2$$
Or :
$${}^{k-1}x\ln\left(\frac{x+y}{2}\right)+{}^{k-1}y\ln\left(\frac{x+y}{2}\right)\geq{}^{k-1}\left(\frac{x+y}{2}\right)*2\ln\left(\frac{x+y}{2}\right)$$
Now we want show that :
$${}^{k-1}x\ln\left(\frac{x+y}{2}\right)+{}^{k-1}y\ln\left(\frac{x+y}{2}\right)\leq {}^{k-1}x\ln\left(x\right)+{}^{k-1}y\ln\left(y\right)$$
Or :
$${}^{k-1}x\ln\left(\frac{\frac{x}{y}+1}{2}\right)+{}^{k-1}y\ln\left(\frac{\frac{y}{x}+1}{2}\right) \ge 0$$
As we have suppose that the function $g(x)=^{2n-1}x$ is convex on the interval $I=[1,\infty)$ (we use Jensen's definition +continuity) next we use Karamata's inequality with $g(x)$ to have the inequalities :
$$x\ln\left(\frac{\frac{x}{y}+1}{2}\right)\ge 0$$
And
$$x\ln\left(\frac{\frac{x}{y}+1}{2}\right)+y\ln\left(\frac{\frac{y}{x}+1}{2}\right) \ge 0$$
It works because the function $g$ is also increasing . Conclude by induction . Obviously log convexity implies convexity also .
We are done
My natural question is :
Is my strategy correct ?
Thanks in advance .
For $x>0$, define recursively $f_n(x)$ by $f_0(x) = 1$ and $f_{n+1}(x) = x^{f_n(x)}$ for every $n \ge 0$. Hence, $f_1(x)=x$, $f_2(x)=x^x$, $f_3(x)=x^{(x^x)}$ and so on.
We prove by recursion that for every $n \ge 0$, $$\forall x \ge 1,~f_n(x) \ge 1 \text{ and } f_n'(x) \ge 0 \text{ and } f_n''(x) \ge 0.$$ When $n=0$, these large inequalities are equalities.
Let $n \ge 0$ and assume that the inequalities above hold. Then for every $x \ge 1$, $$f_{n+1}(x) = \exp(f_n(x)\ln x) \ge \exp(0) = 1,$$ $$f_{n+1}'(x) = \Big(f_n'(x)\ln x + f_n(x)\frac{1}{x} \Big) f_{n+1}(x) \ge (0+1) \times 1 = 1 \ge 0,$$ \begin{eqnarray*} f_{n+1}'(x) &=& \Big(f_n''(x)\ln x + 2f_n'(x)\frac{1}{x} - f_n(x)\frac{1}{x^2} \Big) f_{n+1}(x) + \Big(f_n'(x)\ln x + f_n(x)\frac1x \Big) f_{n+1}'(x) \\ &=& \Big(f_n''(x)\ln x + 2\frac{f_n'(x)}{x} - \frac{f_n(x)}{x^2} \Big) f_{n+1}(x) + \Big(f_n'(x)\ln x + \frac{f_n(x)}{x} \Big)^2 f_{n+1}(x) \\ &=& \Big(f_n''(x)\ln x + 2\frac{f_n'(x)}{x} + \frac{f_n(x)^2-f_n(x)}{x^2} + f_n'(x)^2\ln^2 x + 2f_n'(x)\ln x \frac{f_n(x)}{x} \Big) f_{n+1}(x) \\ &\ge& 0, \end{eqnarray*} which shows the inequalities for the integer $n+1$.
Hence for every $n \ge 0$, the function $f_n$ is non-decreasing and convex on $[1,+\infty[$.