# Inverse second derivative of a Legendre transformation

I'm trying to find the legendre transformation of

$$f(x)=x^3$$

I have calculated it using the approach we learned in class:

1 - Find the derivative of function => $$y(x) = f'(x)$$

2 - Take the inverse of the derivative => $$x(y)$$

3 - (I) Integrate this inverse to find the legendre transformation

3 - or (II) use $$g(y)=-f(x(y))+x(y)y$$

I did this and my Legendre transformation turned out to be

$$g(y)=\frac{2}{3}y\sqrt{\frac{y}{3}}$$

Which I think is correct. However, the next assignment is then

Check the relation $$f''g''=1$$.

Which isn't true:

$$f''=6x$$ and $$g''=\frac{1}{2\sqrt{3y}}$$

I don't understand why this should work. I could not find any information online about a relation between the second derivatives of a function with it's legendre transform.

HINT: Write $$g''$$ in terms of $$x$$, or $$f''$$ in terms of $$y$$.
• Just wanted to add, in general it's a good idea to write out things like $f''$ explicitly if you're running into a wrong answer, since if you'd written it out as $\frac{df^2}{dx^2} \frac{dg^2}{dx^2} = 1$ I think you would've seen the error. – CuriousHegemon Nov 19 '18 at 18:17