Confusion about Legendre transform

Question

So we if we have a function $A(x,y)$ and we want to get a function $B(x,z)$ where $B$ is the function $A$ that is dependent on z and independent of $y$. We define the Legendre transform as $$B(x,z) = zy - A(x,y)$$

and we say that for $B$ to be independent of $y$ then $$z = \frac{\partial A}{\partial y}.$$

I have two questions:

1. We say that $z$ and $y$ are independent variables but how is that true since $\frac{\partial A}{\partial y}$ can depend on $y$?

For example, take the example in Analytical mechanics by Hand & Finch ch.5: $A = y^2(1+x^2)$ thus $ z = \frac{\partial A}{\partial y} = 2y(1+x^2)$ and clearly $z$ depends on $y$,

so how do we say that $z$ and $y$ are truly independent variables?

2. My second question is that in the same book it is said that A(x,y) must be a convex function for the Legendre transform to be defined, but why can it not be defined for concave functions?

As shown in the image attached, this is one geometrical interpretation where we need the function $ux-f(x)$ to be extremized,

but what if the difference can be negative i.e. $f(x) > ux$ such that the difference has a minimum thus $\frac{\partial}{\partial x} (ux-f(x) = 0 $ at a point but this point is now a min. so The maximum does not exist but the minimum does example:

Answer 1

Question 1

The equation \begin{equation} z = \frac{\partial A}{\partial y} \end{equation} is an equation that you are meant to solve to obtain $y(z)$. Then using $y(z)$, we can say that $B(x, z)$ depends only on $x$ and $z$ \begin{equation} B(x,z) = z y(z) - A(x, y(z)) \end{equation} In other words, given $y(z)$ (assuming this function is invertible), it is equivalent to work with $y$ or with $z$ as your variable. The difference is just a change of variables.

Question 2

I believe you are correct. The main property needed is that we can solve \begin{equation} z = \frac{\partial A}{\partial y} \end{equation} for $y(z)$, which means that $\frac{\partial A}{\partial y}$ should depend on $y$, which means that \begin{equation} \frac{\partial^2 A}{\partial y^2} \neq 0 \end{equation} Assuming $A$ is smooth (which we usually do in physics), then this implies that the second derivative of $A$ never changes sign. This condition would apply equally well to convex or concave functions. The main thing that is ruled out are functions that are neither convex nor concave.

I suspect that "convex" is used because typically when the Legendre transform shows up in physics, the function $B$ will be things like the Hamiltonian or the free energy, which we usually to want to minimize.