Consider a function $f$ over $\Re^n$ to $\Re$. Suppose it is true that for every affine subspace with dimension strictly lesser than $n$ the function $f$ is concave. Is the function $f$ concave over $\Re^n$?
To start with, suppose $f: \Re^2 \to \Re$, such that for all $x \in \Re$, $f(x, .)$ is concave and for all $y \in \Re$, $f(., y)$ is concave. Is it possible that $f$ is not concave?
EDIT: Not sure about etiquette here, so I'll leave my original post above. I failed to state the question carefully. Let me give it another try:
Suppose we have real-valued function $f$ on $\Re^n$, and we know the following about the function: Pick a dimension and call it dimension $k$. Next, pick a real number $x$. Now, suppose it is true that $f(., ..., x, ., ...)$ is strictly concave, for any $x$, and for any dimension $k$. Is it true that $f$ is concave?
Theo's response below includes an example $f(x, y), = xy$ that I construe as a counterexample if the qualifier of "strict" were not included.
For what it's worth, this is not a homework question, but instead one born of curiosity.