"Since agent is risk averse, we would expect that $U(E[g]) < U(CE)$ ,
where $CE$ is the certainty equivalent."
This is wrong. I presume the Expected Utility Property holds here, so, if we denote the gamble by $G$, a discrete uniform random variable taking three values according to the setup, we have
$$U(CE) \equiv \sum_{i=1}^3p_iU(g_i) = E[U(G)] < U[E(G)]$$
the inequality to the right due to Jensen's Inequality and the assumption that $U()$ is concave. This also gives us
$$ CE < E(G)$$
which should be intuitive: a risk-neutral person would demand $E(G)$, the expected value of the gamble, in order not to take it. A risk-averse person would require less, to leave the gamble.
Having cleared this, the OP asks: Is it possible that $CE < \min G$?
The answer is : No. Assume that the gamble outcomes are ordered, so $\min G = g_1$.
Ad absurdum, assume that $CE < g_1$ holds. Then we will have
$$U(CE) < U(g_1)$$
Using the definition of $CE$ we replace the left-hand side
$$\sum_{i=1}^3p_iU(g_i) < U(g_1) \implies p_2U(g_2) + p_3U(g_3) < (1-p_1) U(g_1)$$
$$\implies p_2U(g_2) + p_3U(g_3) < p_2U(g_1) + p_3U(g_1)$$
$$\implies p_2[U(g_2)-U(g_1)] + p_3[U(g_3)-U(g_1)] < 0$$
But this is impossible since $g_1 = \min\{g_1,g_2, g_3\}$ and so
$U(g_2)-U(g_1) >0$ and $U(g_3)-U(g_1) >0$.
So assuming $U(CE) < U(g_1)$ led us to an impossible situation, and therefore it cannot hold.
Intuitively, the worst outcome of being in the gamble is to receive the minimum payoff -so for a rational agent, even if it is risk-averse, it would be irrational to accept less than the worst outcome, since then it would certainly do worse than being in the gamble. Note that "aversion to risk" does not mean "take away all risk at all costs".