Example of $g, h:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}$ with this behavior in the origin and this condition on second partial derivatives Last week I had a calculus test consisting in some T/F questions and some open questions. I solved correctly all the TF questions but I got some troubles about two open questions.
The first one was: provide an example of a function $f:\mathbb{R}^3\to\mathbb{R}$ such that
$$\frac{\partial^2 f}{\partial x_1^2} \neq 0 \quad\text{and}\quad \frac{\partial^2 f}{\partial x_1 \partial x_2} =\frac{\partial^2 f}{\partial x_1 \partial x_3}=0.$$
The answer to that first question was easy (I guess, I still do not have the result of the test) because it was enough to take a function $f$ with $\frac{\partial f}{\partial x_1}$ which does not depend on $x_2$ and $x_3$.
The second one was: provide 2 examples of function $g, h:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}$ such that
$$\frac{\partial^2 g}{\partial x_1^2} >0 ,\quad \frac{\partial^2 g}{\partial x_1 \partial x_2} =\frac{\partial^2 g}{\partial x_1 \partial x_3}=0 \quad \quad\text{and}\quad g(x)\le \frac{1}{|x|^2} \text{ for} |x|\le\varepsilon$$
and
$$\frac{\partial^2 h}{\partial x_1^2} >0 ,\quad \frac{\partial^2 h}{\partial x_1 \partial x_2} =\frac{\partial^2 h}{\partial x_1 \partial x_3}=0 \quad \quad\text{and}\quad h(x)\ge \frac{1}{|x|^2} \text{ for} |x|\le\varepsilon,$$
with $\varepsilon$ small enough, i.e. $g, h$ are below/above $\frac{1}{|x|^2}$ in a neighborhood of the origin, respectively.
Here $|x|$ denotes the magnitude of the vector $x=(x_1, x_2, x_3)$.
I was very in trouble with that second question, so much so that I couldn’t answer. I thought about that a lot in these days, but I can’t find an example.
Could someone please provide an example for at least one of the two cases of the second question?
Thank you.
 A: Wont these two choices work ?: $$h(x) = \frac{1}{|x_1|^3},\  g(x) = x_1^2$$
As junjios pointed out, above $h(x)$ is not defined for $x_1=0,x_2 \neq 0$.
A: For $g$ the easiest solution is probably
$$g(x):=x_1^2.$$
Then, if $\vert x\vert\leq 1$ you get
$$g(x)\leq\vert x\vert^2\leq 1\leq\frac{1}{\vert x\vert^2}.$$
Edit: It is not possible to find such a function $g\in C^2\left(\mathbb{R}^3\setminus\{0\}\right)$ which satisfies the estimate
$$g(x)\leq\frac{1}{\vert x\vert^2}$$
everywhere. To see this, define
$$\phi:\mathbb{R}\to\mathbb{R},\ t\mapsto g(t,1,0).$$
Then, by assumption $\phi$ is bounded from above:
$$\phi(t)\leq\frac{1}{1+t^2}\leq 1.$$
But on the other hand, $\phi$ is twice differentiable with
$$\phi''(t)=\partial_1^2g(t,1,0)>0$$
and hence it is strictly convex. Finally, note that a strictly convex (continuously differentiable) function on $\mathbb{R}$ can never be bounded from above (in fact, the only continuously differentiable functions that are both bounded and convex are the constants).
For $h$ I claim that no such sufficiently smooth function exists:
Let's assume for a contradiction that $h\in C^2\left(\mathbb{R}^3\setminus\{0\}\right)$ satisfies
$$\partial_1\partial_2h=0\qquad\text{in }\mathbb{R}^3\setminus\{0\}$$
and
$$h(x)\geq\frac{1}{\vert x\vert^2}\qquad\text{for all }x\in\mathbb{R}^3\setminus\{0\}\text{ with }\vert x\vert\leq\varepsilon.$$
Then, for every $x_3\neq 0$ we have
$$0=\int_0^{\frac{\varepsilon}{2}}\int_0^{\frac{\varepsilon}{2}}\partial_1\partial_2h\left(x_1,x_2,x_3\right)\,\mathrm{d}x_1\mathrm{d}x_2=\int_0^{\frac{\varepsilon}{2}}\left(\partial_2h\left(\frac{\varepsilon}{2},x_2,x_3\right)-\partial_2h\left(0,x_2,x_3\right)\right)\mathrm{d}x_2\\
=h\left(\frac{\varepsilon}{2},\frac{\varepsilon}{2},x_3\right)-h\left(\frac{\varepsilon}{2},0,x_3\right)-h\left(0,\frac{\varepsilon}{2},x_3\right)+h\left(0,0,x_3\right).$$
Now, for all $x_3\neq 0$ that satisfy $\vert x_3\vert\leq\frac{\varepsilon}{2}$ we obtain
$$\frac{1}{x_3^2}\leq h\left(0,0,x_3\right)=h\left(\frac{\varepsilon}{2},0,x_3\right)+h\left(0,\frac{\varepsilon}{2},x_3\right)-h\left(\frac{\varepsilon}{2},\frac{\varepsilon}{2},x_3\right)\\
\leq h\left(\frac{\varepsilon}{2},0,x_3\right)+h\left(0,\frac{\varepsilon}{2},x_3\right)-\frac{1}{\frac{\varepsilon^2}{2}+x_3^2}.$$
As $x_3\to 0$ we observe that the right hand side converges to
$$h\left(\frac{\varepsilon}{2},0,0\right)+h\left(0,\frac{\varepsilon}{2},0\right)-\frac{2}{\varepsilon^2}$$ while the left hand side grows towards infinity. Thus, we've found a contradiction.
