Does separately continuous on compact set imply boundedness? Let $f: \overline{\Omega} \times I \rightarrow \mathbb{R}$, is a continuous function on both variables (separately continuous). $\Omega \in \mathbb{R}^n$ is open, bounded. $I$ is a closed interval in $\mathbb{R}$. 
I don't know if $f$ is bounded on its domain or not. If not, is there any condition to get the boundedness?    
 A: Here's a counterexample.  Take $\overline{\Omega}=I=[0,1]$, and define $f(x,y)=\frac{xy}{x^4+y^4}$ for $(x,y)\neq(0,0)$ and $f(0,0)=0$.  Then $f$ is separately continuous (clearly it is away from $(0,0)$, and if you fix either $x=0$ or $y=0$ then $f$ becomes identically $0$).  But $f$ is unbounded, since $f(1/n,1/n)=n^2/2$ for each $n\in\mathbb{N}$.
I don't know of any natural condition (besides joint continuity) that would guarantee that $f$ is bounded.  As a heuristic reason why you shouldn't expect there to be one, note that separate continuity can be expressed as continuity with respect to a certain topology on the product (the finest topology making the inclusion of each horizontal or vertical line continuous).  This topology is strictly finer than the standard topology, and thus cannot be compact, since no (proper) refinement of a compact Hausdorff topology can be compact.  Given a non-compact topology, it is reasonable to expect that there are continuous $\mathbb{R}$-valued functions that are unbounded (though this is not true for every non-compact space).  So for any condition on $f$ that can be expressed as continuity with respect to some topology finer than the standard topology, we should probably expect it to be possible for $f$ to be unbounded (since the topology is not compact).
