The GBM's distribution at any given point is $$\mathcal N(\mu t,t\sigma^2)$$ Hence the 90th percentile would be $$g(t)=\Phi^{-1}(\mu t,t\sigma^2,0.9)=\mu t +\sqrt t \sigma \Phi(0.9)$$ where $\Phi()$ is the normal or standard normal CDF. This curve is certainly concave and smooth for any $\mu$ and $\sigma^2$. All the convexity comes from the square root of $t$ term.
Consider the second derivative: $$g''(t)=(\mu-\frac 1 2 t^{-1/2}\sigma \Phi(0.9))'=\frac 1 4 t^{-3/2}\Phi(0.9)>0$$
Now, your curve is an empirical 90th percentile, and it's a random curve as such. It will not have a derivative at any point in an ordinary sense, like @whuber mentioned. Hence, in an ordinary sense, i.e. second derivative's sign, it cannot be concave or convex. You'd have to define some measure of convexity for a random curve that doesn't have derivatives in order to judge whether your curve is concave or convex. Obviously, your measure should reconcile with the result I gave you based on theoretical properties of the percentile of GBM. Otherwise, it will not be a sensible measure, if it's not working even in a tractable case where the analytical solution is known.