Bounded second derivative also bounds the function

Question

I've been struggling with this problem for a few days:

Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ defined as $f(x,y)=xg(y)-yg(x)$, where $g:\mathbb{R}\rightarrow \mathbb{R}$ is such that $g\in C^2$, $g(0)=0$ and $|g''(x)|\leq C$ for all $x\in > \mathbb{R}$.

Prove that if $(x,y)\in [0,1]\times[0,1]$, then $|f(x,y)|\leq C$.

I don't know how the second derivative being bounded coul bound the function. The function $f(x,y)=xg(y)-yg(x)$ makes me think of Green's theorem, but that does not get us closer to the second derivative.

$$\int \int_{[0,1]\times[0,1]}xg(y)-yg(x)dxdy=\int_\partial \frac{x^2}{2}g(y)dy+\frac{y^2}{2}g(x)dx$$ I also tried to use Taylor Polynomial with Lagrange Remainder, but we are missing any information about the first derivative of $g$, which appears when calculating the remainder.

I appreciate any help.

Answer 1

By the assumption, $x,y \in [0,1]$. Assume, WLOG, that $x \leq y$. By the Mean Value Theorem, $\exists r,s,t \in (0,1)$ such that $$ \begin{align*} |f(x,y)| &= |xg(y) - yg(x)| \\ &= |xg(y) - xg(0) - yg(x) + yg(0)| \\ &= |x(g(y) - g(0)) - y(g(x) - g(0))| \\ &= |xy g'(ty) - xyg'(sx)| \\ &= xy|g'(ty) - g'(sx)| \\ &= xy|(rty + (1-r)sx)g''(rty + (1-r)sx)| \\ &= [rt xy^2 + (1-r)sx^2y] |g''(rty + (1-r)sx)| \\ &\leq [rtxy^2 + (1-r)sx^2 y] C \\ &\leq C. \end{align*} $$

We apply the Mean Value Theorem thrice:

1. $s$ is applying the Mean Value Theorem to $g$ on the interval $[0, x]$,
2. $t$ is applying the Mean Value Theorem to $g$ on the interval $[0,y]$, and
3. $r$ is applying the Mean Value Theorem to $g'$ on the interval $[sx, ty]$ .

Note the last inequality is true because $x,y,r,s,t \in [0,1]$, by assumption, and $[0,1]$ is closed under multiplication.