Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function with the property that for every \( y \in \mathbb{R} \), the value of the expression \[ \sup_{x \in \mathbb{R}} [xy - f(x)] \] is finite. Define \( g(y) = \sup_{x \in \mathbb{R}} [xy - f(x)] \) for \(y \in \mathbb{R}\). Then

• A. \(g\) is even if \(f\) is even.
• B. \(f\) must satisfy \(\displaystyle \lim_{|x| \to \infty} \frac{f(x)}{|x|} = +\infty.\)
• C. \(g\) is odd if \(f\) is even.
• D. \(f\) must satisfy \(\displaystyle \lim_{|x| \to \infty} \frac{f(x)}{|x|} = -\infty.\)

Hint:

For a Legendre transform to be finite everywhere, the original function must grow faster than any linear function — i.e., \(\frac{f(x)}{|x|} \to +\infty.\)

Answer 1

The Correct Option is A, B

Step 1: Interpretation of the expression.
The function \(g(y)\) defined by \(g(y) = \sup_{x \in \mathbb{R}} [xy - f(x)]\) is the \textit{Legendre transform} (or convex conjugate) of \(f\). For the supremum to be finite for all \(y\), the linear function \(xy\) must eventually be dominated by \(f(x)\) as \(|x| \to \infty.\)
Step 2: Growth condition.
If \(f(x)\) grows slower than linearly, say \(\frac{f(x)}{|x|}\) does not tend to \(+\infty\), then for some \(y\), \(xy - f(x)\) could become arbitrarily large, making the supremum infinite. Thus, we require \[ \lim_{|x| \to \infty} \frac{f(x)}{|x|} = +\infty. \]
Step 3: Conclusion.
Hence, (B) is correct.