What does this mean in the answer by Qmechanic: $$f|_U~=~q^1 \quad \text{and}\quad g|_U~=~p_1 $$

@ACuriousMind ah, yes. Thanks!

$f$ and $g$ will be canonical conjugates as long as $$\exists \:\:\: i\in\{1,\dots,n\} \quad;\quad f|_U~=~q^i \quad \text{and}\quad g|_U~=~p_i $$ Am I right?

@ACuriousMind as long as there exists a certain $i$ for which the expression holds, that's what I meant.

@ACuriousMind But what would stop me from hypothesizing $$f|_U~=~q^2 \quad \text{and}\quad g|_U~=~p_2$$ and thus calling them conjugate variables? I am only interested in confirming that the indices are immaterial here, aren't they?

@ACuriousMind right, thanks! Also, can't we just directly define $f$ and $g$ to be conjugate variables by saying that $$\{f,g\}=1$$ where $\{\}$ is the poisson bracket?

@ACuriousMind yeah, that's what I was trying to confirm. It's easy foor a noob like me to misunderstand obvious things :-)

@ACuriousMind But can't we just use the definition of the Poisson bracket? Can we at least do it in the cases where $f=q^i$ and $g=p_i$ over the whole $\mathbb R^{2n}$ space?

@ACuriousMind Alright, I think I am getting ahead of myself now, and I gotta stop before I misunderstand stuff. Thanks a lot for your help :-))))

FWIW, I was able to make a (satisfactory) sense of Legendre transforms. Thanks guys.

@bolbteppa So, then, I do expect them to follow the properties that canonical coordinates follow, namely $\{f,g\}=1$. Bht ACM says otherwise and I am too unexperienced to understand why.

@bolbteppa I was thinking about it in the wrong way. I was thinking of the transform as some sort of rewriting the function so that it stays the same, or has same characteristics (akin to rewriting a function in terms of the sines and cosines of the constituent frequencies, like we do in fourier decomposition). But I later realized that we are willingly defining a different and new function, with different properties and thus, different equations (Hamiltonian EoM different from Lagrangian EoM).

It took a while for me to understand how both the functions are different yet completely equivalent from a physical POV.

@ACuriousMind Ah, I see. The age old misunderstanding of "if $A$ is true, then $B$ gotta be true, but if $B$'s true, then $A$ need not be true". In our case $A\equiv$ $f$ and $g$ being conjugate variables; and $B\equiv \{f,g\}=1$, right?

@bolbteppa Yeah, but defining a new function just felt a bit out-of-the-thin-air kind of thing to me, until I realized why are we doing it the way we are doing it.

I am current!y experiencing pure satisfaction, knowing that I finally understand a tiny bit of this stuff which I struggled with for the past 3 days. Thanks again!