Legendre Transform of C1 function

Question

If $f:R \to R$ is a $C^1$ and strictly convex function with superlinear growth (i.e $ \lim_{|p|\to \infty}f(p)/|p| = \infty $),then what can we say about $f^*$ ?. will it be $C^1$ ?

Answer 1

Hint: $f^*(p) = \max_x \{px - f(x)\}$ so $p=f'(x)$ and $$f^*(p) = p(f')^{-1}(p) - f((f')^{-1}(p)).$$ [Note $(f')^{-1}$ exists because $f$ is strictly convex, so $f'$ is strictly increasing.]

Here are a few more details: Write $g=(f')^{-1}$ so that $f^*(p) = pg(p)-f(g(p))$. Then

$$f^*(p+h)-f^*(p) = hg(p+h) + p(g(p+h)-g(p)) - (f(g(p+h))-f(g(p)) ).$$

By a Taylor expansion

$$f(g(p+h))-f(g(p)) = f'(g(p))(g(p+h) - g(p)) + o(h)= p(g(p+h)-g(p)) + o(h).$$

Plugging this in above we get

$$f^*(p+h) - f^*(p) = hg(p+h) + o(h).$$

Dividing by $h$ and sending $h\to 0$ we get that $(f^*)'$ exists and $(f^*)'(p) = g(p)$, so $f^*$ is $C^1$.