User blog comment:Edwin Shade/A Proof/@comment-1605058-20171104122735/@comment-1605058-20171109202620

I did not intend this function to be a contender for the "real tetration", but rather just as some extension of the integer tetration to all positive real numbers.

Either way, we can actually circumvent that. Let me fix \(a=e\) for simplicity, then consider a function \(f:[0,1]\to\mathbb R\), infinitely differentiable, which at \(0\) has a Taylor expansion \(1+x\), whereas the Taylor expansion at \(1\) is the same as one for \(e^x\). I won't dig into details of how this function can be constructed, but it is possible and we may even take \(f\) to have all its derivatives increasing (so that it is increasing and very smoothly so). This being done we can define \(e\uparrow\uparrow b=f(b)\) for \(b\in [0,1]\) and \(e\uparrow\uparrow(b+1)=e^{e\uparrow\uparrow b}\) for \(b>1\). The choice of \(f\) guarantees this function will be infinitely differentiable and smoothly increasing.