Here's a fun counterexample: a function $\mathbb{R}^n \to \mathbb{R}$ that is jointly convex in any $n-1$ of the variables, but not in all variables at once. The function is

$f(x_1,\ldots,x_n) = \frac{1}{2}(n-1.5)\sum_{i=1}^n x_i^2 - \sum_{i < j} x_ix_j$

To see why this is, note that the Hessian of $f$ is equal to

$\left[ \begin{array}{cccc} n-1.5 & -1 & \cdots & -1 \ -1 & n-1.5 & \cdots & -1 \ \vdots & \vdots & \ddots & \vdots \ -1 & -1 & \cdots & n-1.5 \end{array} \right]$

This matrix is equal to $(n-0.5)I - J$, where $I$ is the identity matrix and $J$ is the all-ones matrix, which is rank 1 and whose single non-zero eigenvalue is $n$. Therefore, this matrix has $n-1$ eigenvalues of $n-0.5$, as well as a single eigenvalue of $-0.5$, and hence is not positive definite.

On the other hand, any submatrix of size $n-1$ is of the form $(n-0.5)I-J$, but where now $J$ is only $(n-1) \times (n-1)$. This matrix now has $n-2$ eigenvalues of $n-0.5$, together with a single eigenvalue of $0.5$, and hence is positive definite. Therefore, the Hessian is positive definite when restricted to any $n-1$ variables, and hence $f$ is convex in any $n-1$ variables, but not in all $n$ variables jointly.

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