Is "$f(k,l)$ is decreasing return to scale $\Leftrightarrow f_{ll}f_{kk}-f_{kl}^2>0$" always true? For the productions $f(k,l) $ that are continuously differentiable, is the proposition  that

"$f(k,l)$ is decreasing return to scale $\Leftrightarrow f_{ll}f_{kk}-f_{kl}^2&gt;0$"

always true, I have checked for Cobb-Douglas functions and believe it will also hold for all Constant-Elasticity fuction, can someone give me some hints or show some counter-examples?

Here we define $f(k,l)$ to be decreasing return to scale if $ f(tk,tl)&lt;tf(k,l)$ for $t&gt;1$ for all $(k,l)$ in the domain.
In general the statement is wrong. Here is a counterexample:

Suppose you have $f(k,l) = -k l^\beta$ with $\beta &gt;0$ and $(k,l)\in\mathbb{R}^2_{++}$ (you can interpret $f$ as a production function for a "bad" commodity). Then you have:
$$f(tk,tl) = - t^{1+\beta} k l^\beta = t^{1+\beta} f(k,l) &lt; t f(k,l)$$
so that $f(k,l)$ is decreasing returns to scale. 

Now let's evaluate $ f_{ll}f_{kk} - f^2_{kl} $. Since $f$ is linear in $k$ we have that $f_{kk}$ is zero, so it is $ f_{ll}f_{kk}$.  On the other hand, we have $f_{kl} = -\beta l^{\beta-1}$ so that overall we have:
\begin{equation}
f_{ll}f_{kk} - f^2_{kl} = 0 - \left[ -\beta l^{\beta-1} \right]^2 = - \beta^2 l^{2(\beta -1)} &lt;0
\end{equation}

So then, where is the trick?

The fact is that $ f_{ll}f_{kk} - f^2_{kl} &gt; 0$ is not a sufficient condition to conclude that the hessian of $f(k,l)$ is negative semi-definite and so $f$ is concave because you also have to impose restrictions on the sign of the principal minor of order 1 of the hessian matrix.

However, provided the hessian matrix is indeed negative semi-defintie than you conclude $f$ is concave and decreasing returns to scale will follow, the statement then holds.