A convexity question Let $Q=[0,1]\times[0,1]$ and let $a$ be a positive smooth function on $Q$. Does there exist a smooth positive function $u$ On $Q$ such that there holds
$$ \frac{\partial^2}{\partial x_1^2}u <0 $$
and
$$ \frac{\partial^2}{\partial x_2^2}(au)<0 $$
where both inequalities are to be satisfied everywhere on $Q$.
 A: This is not always possible. Note that if $f(x)$ is a non-negative concave function on a segment $\Delta$, and $\alpha,\beta,2\beta-\alpha$ belong to $\Delta$, then $f(\beta)\geqslant (f(\alpha)+f(2\beta-\alpha))/2\geqslant f(\alpha)/2$.
Now choose a square $ABCD$ with horizontal side $AB$ which is close enough to the centre of $Q$ (say, coordinates of all vertices $A,B,C,D$ are between $0.49$ and $0.51$). Then by our observation the values $f(A)$, $f(B)$ have a ratio between $1/2$ and $2$, so are $f(C)$ and $f(D)$; $a(A)f(A)$ and $a(D)f(D)$, $a(B)f(B)$ and $a(C)f(C)$. This is not possible if $a(A)=a(C)=1$, $a(B)=a(D)=100$.
A: Continuous (and therefore especially smooth) real-valued functions on a compact set like $Q$ are bounded, therefore there is a constant $C_1$ so that $\partial_2^2a<C_1$, a constant $C_2$ so that $|\partial_2a|<C_2/4$ and a constant $C_3$ so that $a>C_3$. The constant $C_3$ can be chosen to be positive as $a(Q)$ is a compact subset of $\mathbb{R}^+$, which is open.
Take $u(x_1,x_2)=b(1-x_1^2)+c(1-x_2^2)+1$ with $b,c>0$ and $0< u(x_1,x_2)\leq b+c+1$, then:
\begin{equation}
\partial_1^2u(x_1,x_2)=-2b<0
\end{equation}
\begin{equation}
|\partial_2u(x_1,x_2)|=|-2cx_2|\leq 2c
\end{equation}
\begin{equation}
\partial_2^2u(x_1,x_2)=-2c<0.
\end{equation}
Therefore, using that $u$ is positive for the first term, we have:
\begin{align*}
\partial_2^2(au)
&=(\partial_2^2a)u+2(\partial_2a)(\partial_2u)+a(\partial_2^2u)
<C_1u+C_2c+C_3(\partial_2^2u) \\
&\leq C_1(b+c+1)+C_2c-2C_3c
=C_1(b+1)-(2C_3-C_2-C_1)c.
\end{align*}
Now we need additional conditions for the constants ($C_1<0$ or $C_1+C_2<2C_3$) both translating into necessary conditions ($\partial_2^2a<0$ or $\partial_2^2a+4|\partial_2a|<2a$) for $a$ that are not given.
I unfortnutly only have part of a possible solution here.
