We say that a function $f:A→A$ is bounded if: There exist a positive real number $M $ such that $$|f(z)|≤M$$ for all $z$ in the set $A$.
My question is: Is the Dirichlet eta function (https://en.wikipedia.org/wiki/Dirichlet_eta_function) is bounded in the sense of the above defition. Here $A$ is the set in the complex plane with $Re(z)>0$.
As is written in the wikipedia page you linked, the Dirichlet eta function can be extended to an entire function, i.e. a function that is holomorphic on the whole complex plane. But by Liouville's theorem, such a function, if it is bounded, is necessarily constant, which is clearly not the case.
Edit: as pointed out in the comments this answer is not appropriate because I considered the boundedness of the extension $\tilde{\eta}$ of the Dirichlet eta function (with domain $\mathbb{C}$) rather than the Dirichlet eta function $\eta$ defined on $\Re (z)>0$.
However, suppose by contradiction that $\eta$ is bounded, i.e. $\tilde{\eta}$ is bounded for $\Re(s)>0$, with $|\tilde{\eta}(s)|=|\eta(s)|\leq M$. The relation (which you can also find in the wikipedia page) $$\tilde{\eta}(-s)=2\frac{1-2^{-s-1}}{1-2^{-s}}\pi^{-s-1}s\cdot \sin\left(\frac{\pi s}{2} \right)\Gamma(s)\eta(s+1)$$ holds for all $s$ with $\Re(s)\geq 0$ (notice that in this case, $\Re(-s)\leq 0$, and $\Re(s+1)>0$). Moreover, the term in the right-hand-side is bounded for $\Re(s)\geq 0$ and therefore for some $M'>0$ we have $$|\tilde{\eta}(-s)|\leq M',\qquad \forall s:\Re(s)\geq 0$$ and hence $$ |\tilde{\eta}(s)|\leq \max\left\{M;M'\right\},\qquad \forall s\in \mathbb{C}$$ i.e. the entire function $\tilde{\eta}$ is bounded and non-constant, contradicting Liouville's theorem.
EDIT: The answer is still wrong. Even if $\eta(s+1)$ is bounded, the term in the right hand side is unbounded for $\Re(s)>0$ due to $\Gamma(s)$ growing faster than an exponential as $\Re(s)\to +\infty$.
I will delete this answer as soon as @reuns posts his own - which for now is sketched in the comments below this answer.