assume $\mathcal X$ is a convex set and $\mathit f(x,a)$ is a function that $$f:\mathcal X \times \Bbb R \to \Bbb R$$
$f(x,a)$ is convex with respect to $x$ for every $a$ and $f(x,a)$ is convex with respect to $a$ for every $x$
Is that function $f(x,a)$ is convex with respect to $(x,a)$??
if not, under what condition that can provide convexity to guarantee function $f(x,a)$ is convex with respect to $(x,a)$?
I know it goes to prove $f((1-\lambda )(x,a)+\lambda(x',a')) \le (1-\lambda)f(x,a)+\lambda f(x',a') $
but I didn't know how to go on when I take this step
$f((1-\lambda )(x,a)+\lambda(x',a'))$ = $f((1-\lambda)x+\lambda x' ,(1-\lambda)a+\lambda a')$
No, in order to guarantee concavity in more than one dimension, you need the Hessian Matrix to be positive semi definite over the entire domain. For an example of a function that is convex in every independent variable, but not in the combined domain look at
$$ xy - x - y $$
on $\mathbb{R}x\mathbb{R}$. This function is linear with respect to each independent variable, however it is not convex as a whole. There is even a saddle point at $(1,1)$