I am referring to the ISS definition by Sontag of
${\displaystyle |x(t)|\leq \beta (|x_{0}|,t)+\gamma (\|u\|_{\infty }).}$
I understand that 0-GAS is a necessary condition for ISS. But is GAS for all constant u a sufficient condition?
Update:
For clarification, I mean the system is GAS for an arbitrary, but constant input u.
As of my understanding now, BIBO stability and 0-GAS are necessary for ISS. So the upper condition cannot be sufficient. So the updated question is:
Is a system, that is BIBO stable and GAS for an arbitraty, but constant input automatically ISS?
I think this is not the case.
First, to state the hypothesis from your question: Given $$ \dot{x}=f(x,u)\tag{1} $$ such that $(1)$ is globally asymptotically stable (GAS) for all $u\in \mathbb{R}^m$. Then $(1)$ is input to state stable (ISS).
I claim this is wrong by reusing a well known counterexample from linear time varying (LTV) system analysis: $$ \begin{align} \dot{x}_1 &= -x_1 - 5 x_2 \cos(u)^2 + 2.5 x_1 \sin(2 u) \\ \dot{x}_2 &= -x_2 + 5 x_1 \sin(u)^2 - 2.5 x_2 \sin(2 u) \end{align}\tag{2} $$
For any $u\in\mathbb{R}$, the system $(2)$ has both its eigenvalues at $-1$, so it is GAS for any (finite) constant $u$. However, take the bounded input $$ u(t)=2\pi\left(\frac{t}{2\pi} - \left\lfloor\frac{t}{2\pi}\right\rfloor\right) \tag{3} $$ where $\lfloor\cdot\rfloor$ is the floor function. With this input, the system $(2)$ is equivalent to the first example of the 2.2 Examples section from
Ilchmann, A., Owens, D. H., & Prätzel-Wolters, D. (1987). Sufficient conditions for stability of linear time-varying systems. Systems & control letters, 9(2), 157-163.
Note that they cite another paper as the original source for this system, which is page 3 of
Coppel, W. A. (1978) Dichotomies in Stability Theory. Lecture Notes in Mathematics, Vol. 629, Springer-Verlag, Berlin.
There, the system is cited as an example of an LTV system that is unstable although for every single $t$ its $A(t)$ matrix is Hurwitz. In the paper they use $u(t)=t$, since they investigate LTV systems, but since the time variable appears only in the trigonometric sine/cosine functions, we can redefine it as a bounded input by $(3)$ and end up with the same dynamical system since $(3)$ is just a sawtooth function with $2\pi$-period.
So although $(2)$ is GAS for every constant $u\in\mathbb{R}$, it is not ISS because it becomes unstable under the bounded input $(3)$.
Edit: Addressing your updated question. Even if you you additionally assume that your system is bounded-input, bounded-output (BIBO) stable, that doesn't help anything in asserting whether the system is ISS or not.
Consider the system $(2)$ and add a new state equation: $$ \begin{align} \dot{x}_1 &= -x_1 - 5 x_2 \cos(u)^2 + 2.5 x_1 \sin(2 u) \\ \dot{x}_2 &= -x_2 + 5 x_1 \sin(u)^2 - 2.5 x_2 \sin(2 u) \\ \dot{x}_3 &= -x_3 \end{align} $$ with output $y=x_3$. Clearly, this new system is BIBO stable. It is also GAS for every constant $u$ (like before). However, it is not ISS, because the bounded input $(3)$ still makes the norm of the state vector diverge to infinity.