Proving that $|x \, g(y) - y \, g(x)| \leq C$ for $x,y \in [-1,1]$, where $|g''(t)| \leq C$ Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $f(x,y) =  x \, g(y) - y \, g(x)$, where $g:\mathbb{R} \rightarrow \mathbb{R}$ such that $g \in C^2$, $g(0)=0$ and $|g''(x)| \leq C $ $\forall x \in \mathbb{R}$. I Want to prove:
a) If $(x,y) \in [-1,1]\times[-1,1]$ then $|f(x,y)| \leq C$.
b) Evaluate $\int^1_0 \int^1_0 f(x,y) \, dx \, dy$.
For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.
But, for the a part I'm getting trouble.
I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|\leq$ sup $ g^{''}$ $|x|$ on $x \in [-1,1]$ and then conclude $ |g'(x)| \leq C$ either.
So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.
 A: a) If we change from $g$ to $\widetilde{g}(x) = g(x) + a x$ for any $a \in \Bbb{R}$, this does not change the function $f$. Also, we will still have $\widetilde{g}(0) = 0$ and $|\widetilde{g}''(x)| \leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.
Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x \in [-1,1]$, there is some $\xi \in [0,x] \cup [x,0] \subset [-1,1]$ satisfying
$$
g(x) = g(0) + g'(0) \cdot x + \frac{g''(\xi)}{2} (x - 0)^2 \, ,
$$
and hence $|g(x)| \leq \frac{C}{2} \cdot x^2 \leq \frac{C}{2}$ for all $x \in [-1,1]$.
Now, we easily see
$$
|f(x,y)| \leq |x| \cdot |g(y)| + |y| \cdot |g(x)| \leq \frac{C}{2} + \frac{C}{2} = C
$$
for all $x,y \in [-1,1]$.
Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.
b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := \int_0^1 t \, d t$ (so that $a = 1/2$) and $b := \int_0^1 g(t) \, dt$, then
\begin{align*}
\int_0^1 \int_0^1 f(x,y) \, d x \, d y
& = \int_0^1 \int_0^1 x \, g(y) \, d x \, d y
    - \int_0^1 \int_0^1 y \, g(x) \, d x \, d y \\
& = \int_0^1 g(y) \cdot a \, d y
  - \int_0^1 y \, b \, d y \\
& = a \cdot b - a \cdot b
  = 0 \, .
\end{align*}
