Showing a composition is in $L^2$

Question

Let $f$ be a smooth increasing function passing through the origin with $|f(x)| \leq C_1(1+|x|)$. Define $F(r) = \int_0^r f(s)\;ds$ and $F^*(t) = \sup_{r \in \mathbb{R}}tr - F(r).$ Note that $F^*(f(x)) = xf(x) - F(x)$.

If $u \in L^2(\Omega)$ for $\Omega$ bounded, then $F^*(f(u)) \in L^2(\Omega)$.

How do I prove this? I can't show that $F(u) \in L^2$, only that it is in $L^1$ using the growth condition.

Answer 1

In general, this is not true. Consider

$$f(x) := x,$$

then $F(x) = \frac{x^2}{2}$, and therefore

$$F^{\ast}(f(u)) = u \cdot f(u) - F(u) = u \cdot u - \frac{u^2}{2} = \frac{u^2}{2}.$$

This shows that $F^{\ast}(f(u)) \in L^1$, but not necessarily $F^{\ast}(f(u)) \in L^2$.