lagrangian formalism - Why are Hamiltonian Mechanics well-defined?

I have encountered a problem while re-reading the formalism of Hamiltonian mechanics, and it lies in a very simple remark.

Indeed, if I am not mistaken, when we want to do mechanics using the Hamiltonian instead of the Lagrangian, we perform a Legendre tranformation on the Lagrangian to get the Hamiltonian. This, in the case of a 1 dimensional problem, is written as follows: $$ H(p,q) = p\dot{q}-L(q,\dot{q}).$$ Notice that this transformation is such that $H=H(p,q)$, where $p = \frac{\partial L}{\partial \dot{q}}$. By variation of $H$, we can indeed verify it as function of $p$ and $q$, so that they are now considered the new independent variables of the problem.

So far so good. However, there is a problem, and it lies in the fact that for this construction to hold, we need the Lagrangian not to change convexity. Let me write what I know about the Legendre transformation, in a somewhat formal way:

Given a function $f: x\rightarrow f(x)$, we define the function $p: x\rightarrow p(x)$ by the relation $\frac{df}{dx} = p$. Supposing $\frac{df}{dx}$ is invertible, we can define the inverse of $p(x)$, which we call $g : p\rightarrow g(p)$. Then, the legendre transformation of $f$ is $f^* : p\rightarrow f^*(p)$ such that $\frac{df^*}{dp}=g(p)$. We can write, in a more familiar way, $g(p) = x(p)$ since it is the inverse of $p(x)$.

Anyway, we can prove with those assumption that $f^* = pg(p)-f(g(p))$ which is $f^* = px-f(x)$ written in the "familiar" way. All that is just to point out that, for all this construction to work, and hence for the existence of $f^*$, we need the condition that $f(x)$ be of constant convexity, otherwise $\frac{df}{dx}$ is not invertible and we cannot even define $g(p)$.

However, when we consider a general Lagrangian, I don't think that this is always the case. Taking simply $L = \dot{q}^3$ makes the Lagrangian not of constant convexity. And yet, we always use the Hamiltonian, without ever checking this convexity constraint. Why can we do this? Is it because we are interested in the local behavior of our Lagrangian? But even then, what would we do at an inflexion point?

Or is it because a general "physical" Lagrangian will always satisfy the condition of constant convexity?