About boundness of a function
Consider a function $f\in C^2(\mathbb{R},\mathbb{R})$ whose first and second derivatives are globally bounded and such that $f(x,0)=f(0,y)=0$ for all $(x,y)\in \mathbb{R}^2$.
I should prove that there exists $C>0$ s.t. $|f(x,y)|\leq C|xy| $ for all $(x,y)\in \mathbb{R}^2$ .
I have thought to prove that the function $\frac{f(x,y)}{xy}$ is bounded but I actually couldn't argue it. Any suggestion?
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$\begingroup$Hint/Start: Of course the function is bounded for finite $(x,y)$ so we need to show that it stays bounded on the axes and for $x,y\rightarrow \infty$.
Basically this is equivalent to saying $\frac{f(x,y)}{xy}$ is continuous outside the axes, so if it diverges it must diverge on the boundary of this domain, i.e. at one of the above places.
On the axes:
Note that for any fixed $x,y\neq 0$: $$\frac{f(x,y)}{xy}=\frac{f(x,y)-f(x,0)}{xy}=\frac{f(x,y)-f(0,y)}{xy}.$$ Thus as $x\rightarrow 0$ $y\neq 0$ $$\lim_{x \rightarrow 0}\frac{f(x,y)}{xy}=\lim_{x \rightarrow 0} \frac{f(x,y)-f(0,y)}{x}\cdot\frac{1}{y}=\frac{\partial_xf(0,y)}{y},$$ and this is bounded by assumption on $f$.
In a similar way we can bound the term on the $O_x$ axis away from the origin.
The remaining problem is for $x,y$ both going to zero, i.e. at the origin, and here is where the second derivative kicks in.
Can you formalize this last step? It can be done in a very similar way to what I did above, using that $\partial_xf(0,0)=0=\partial_yf(0,0)$, which is a consequence of $f$ being constant on the axes.
For $x,y\rightarrow \infty$:
Similarly to the previous argumentation and using the mean value theorem: $${f(x,y)}={f(x,y)-f(x,0)}=|y|\partial_yf(x,y^\ast),y^\ast \in[0,y].$$ Thus $$\left|\frac{f(x,y)}{xy}\right|\leq \frac{|y|\|\partial_yf\|_{C^0}}{|x||y|}\leq C\frac{1}{|x|},$$ so the function is uniformly bounded on the whole plane.
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