If supp$(g)\subset[a,b]$ for some $a<b\in\Bbb R$ and $g\not\equiv 0$ then what can we say about the smoothness of $g$?
If $\exists \{x_n\}_{n\geq 1}\text{a monotone sequence such that}\ g(x_n)\ne0\ \forall n\geq1$ then surely the sequence must converge because it is monotone and bounded by $a$ from below and $b$ from above.
Nothing prevents $g$ from being continuous but what about $C^1$ or higher $C^k$?
Now by adding an extra condition:
So let's assume $g$ is continuous and $\exists \tilde x\in(a,b)$ such that $g'(x)\ge0\forall x\le \tilde x$ and $g'(x)\le0\forall x\ge \tilde x$
Now can $g$ ever be $C^k$ if $k>0$ and up to what $k$?
My intuition tells me that there must be a point for which the slope changes abruptly from $0$ to something $>0$ soone after $a$ and from something $<0$ to $0$ soon before $b$... But I don't know.
$\inf\{x\le b\ :\ g(x)=0\}=:x_0\in\Bbb R_{\le b}$ exists because the set is non empty and it would be a discontinuity point of $g'$
Then on the other hand if I look at the function 
that is $0$ if $x<0$ and $1-\sqrt{1-x^2}$ if $x\ge0$ I kind of want to take back those arguments
It is always possible to construct a $g$ having these properties and such that $g\in C^{\infty}(\mathbb{R})$. For simplicity, let $[a,b]=[-1,1]$. Then the function $$g(x):=\begin{cases}e^{1/(x^2-1)} & |x|\leq 1 \\ 0& |x|\geq 1\end{cases}$$ is $C^{\infty}(\mathbb{R})$, $g\neq 0$, and $g=0$ outside of $[-1,1]$. (to see this, prove by induction on $n$ that $g^{(n)}(x)=\frac{p_n(x)}{q_n(x)}e^{-1/(1-x^2)}$ for some polynomials $p_n,q_n$. Since the exponential decays faster at $\pm 1$ than any polynomial, the $n$-th derivative tends to $0$ as $x\to \pm 1$).
However, such a function is necessarily not analytic, because its Taylor series at $\pm 1$ is equal to $0$, yet $g\neq 0$ (so it also provides an example of a smooth non analytic function).
Regarding your second condition, the example provided is such that $g'(x)\geq 0$ for $x\leq 0$ and $g'(x)\geq 0$ for $x\geq 0$.