is $f(x,y)=e^{x-y}$ globallyconvex? The function is positive $\forall (x,y)\in R^2$.If I consider the restriction to y-axis $f(0,y)=e^{-y}\rightarrow 0^+$ for $y\rightarrow +\infty$ so Inf f=$0$.
To study convexity can I consider the matrix Hessian?
 A: For convexity I would start by changing coordiantes from $(x,y)$ to $(p,q)=(x-y,x+y)$. Then $f(p,q)=e^p$, and it is easy to directly compare $f(v)$ and $f(w)$ with $f(tv+(1-t)w)$. One then sees that the function inherits convexity from the real exponential function.
A: If $p = (p_1,p_2)$ and $q = (q_1,q_2)$ and $0 \le t \le 1$, 
$$f(t p + (1-t) q) = \exp(t(p_1 - p_2)+(1-t)(q_1-q_2)) \le
t \exp(p_1-p_2) + (1-t) \exp(q_1 - q_2) = t f(p) + (1-t) f(q)$$
by convexity of $\exp$.
Somewhat more generally, if $g$ is an affine map and $h$ is convex then $h \circ g$ is convex.
A: Take any $(x,y), (z,w)\in\mathbb{R}^2$, and any $t \in \mathbb{R}$. Then $f(t(x,y)+(1-t)(z,w)) = e^{tx+(1-t)z - ty - (1-t)w} = e^{t(x-y)+(1-t)(z-w)}$, and $tf(x,y)+(1-t)f(z,w) = te^{x-y}+(1-t)e^{z-w}$. We need to show that the former is never greater than the latter. 
But $g: \mathbb{R}\to\mathbb{R}: x \mapsto e^x$ is convex, so we have (taking our points in the definition of the convexity of $g$ to be $x-y$ and $z-w$) 
$$e^{t(x-y)+(1-t)(z-w)} \leq te^{x-y}+(1-t)e^{z-w},$$
But the left-hand-side of this is exactly $f(t(x,y)+(1-t)(z,w))$, and the right-hand-side is exactly $tf(x,y)+(1-t)f(z,w)$, so indeed, $f$ is convex. 
