Let $f:\mathbb{R}^n\to\mathbb{R}$ be a function such that it is convex in each variable (with all another fixed). Prove that $f$ is continuous.
Suppose $(x_2,\ldots,x_n)$ are fixed. As a function of $x_1$, $f$ has left and right derivatives everywhere. I suspect these derivatives are actually equal (so $f$ is differentiable as a function of $x_1$). I don't think anything can be said in terms of second-order derivatives.
One probably has to prove that $f$ is continuous from first principles (mimicking the proof for usual convex functions).
This proof is adapted from the proof of Theorem 2.31 in Dacorogna's Direct Methods in the Calculus of Variations. It is quite similar to the usual proof of continuity of convex functions.
Let $x_0\in \mathbb R^n$. Let us prove first that $f$ is bounded above in any neighborhood of $x_0$.
WLOG we may assume $x_0=0$. Consider $\epsilon >0$ and let $a=\max \{f(x),\; \forall i, x_i\in\{-\epsilon, \epsilon \}\}$ be the maximum of $f$ over the vertices of the cube centered at $0$ with side length $2\epsilon$. Let us prove that $\|x\|_\infty \leq \epsilon \implies f(x)\leq a$.
Let $\epsilon_1,\ldots,\epsilon_{n-1}\in \{-\epsilon, \epsilon \}^{n-1}$ be arbitrary. Separate convexity w.r.t $x_n$ yields $$\begin{aligned} \forall x_n \in [-\epsilon, \epsilon], f(\epsilon_1,\ldots,\epsilon_{n-1}, x_n) &\leq \frac{\epsilon-x_n}{2\epsilon}f(\epsilon_1,\ldots,\epsilon_{n-1}, -\epsilon)+ \frac{\epsilon+x_n}{2\epsilon}f(\epsilon,\ldots,\epsilon_{n-1}, \epsilon)\\ &\leq \frac{\epsilon-x_n}{2\epsilon}a + \frac{\epsilon+x_n}{2\epsilon} a \\ &=a \end{aligned}$$ For arbitrary $\epsilon_1,\ldots,\epsilon_{n-2}\in \{-\epsilon, \epsilon\}^{n-2}$ and any $x_{n} \in [-\epsilon, \epsilon]$, separate convexity w.r.t $x_{n-1}$ yields $$\begin{aligned} \forall x_{n-1} \in [-\epsilon, \epsilon], f(\epsilon_1,\ldots,\epsilon_{n-2}, x_{n-1}, x_n) &\leq \frac{\epsilon-x_{n-1}}{2\epsilon}f(\epsilon_1,\ldots,\epsilon_{n-2}, -\epsilon, x_n)+ \frac{\epsilon+x_{n-1}}{2\epsilon}f(\epsilon_1,\ldots,\epsilon_{n-2}, \epsilon, x_n)\\ &\leq \frac{\epsilon-x_{n-1}}{2\epsilon}a + \frac{\epsilon+x_{n-1}}{2\epsilon} a \\ &=a \end{aligned}$$ Iterating this process yields the claim.
Let us prove that $f$ is continuous at $0$. By the claim, there is some $a$ such that $\|x\|_\infty \leq 1 \implies f(x)\leq a$.
Let $\epsilon>0$. The upper bound $a$ can be chosen so that $\epsilon \leq an2^n$ holds. Set $\delta = \frac{\epsilon}{an2^n}\leq 1$. Let us prove that $$\|x\|_\infty \leq \delta \implies |f(x)-f(0)|\leq \epsilon$$ $\bullet$ Writing $x_1=\delta \frac{x_1} \delta + (1-\delta)0$ then $x_2=\delta \frac{x_2} \delta + (1-\delta)0$, etc... iteratively yields $$f(x)\leq \sum_{k=1}^n \delta (1-\delta)^{k-1}f(0,\ldots,0,\frac{x_k}\delta,x_{k+1},\ldots,x_n) + (1-\delta)^nf(0)$$ Since $\sum_{k=1}^n \delta (1-\delta)^{k-1} = 1-(1-\delta)^{n-1}$, one gets $$\begin{aligned} f(x)-f(0) &\leq \sum_{k=1}^n \delta (1-\delta)^{k-1} \left(f(0,\ldots,0,\frac{x_k}\delta,x_{k+1},\ldots,x_n) - f(0) \right)\\ &\leq a \sum_{k=1}^n \delta (1-\delta)^{k-1} \\ &\leq a \delta n \leq \frac{\epsilon}{2^n}\\ &\leq \epsilon \end{aligned}$$
$\bullet$ Writing $0=\frac{1}{1+\delta} x_1 + \frac{\delta}{1+\delta}\left(-\frac{x_1}\delta\right)$ then $0=\frac{1}{1+\delta} x_2 + \frac{\delta}{1+\delta}\left(-\frac{x_2}\delta\right)$, etc... iteratively yields $$f(0)\leq \frac{1}{(1+\delta)^n}f(x)+\delta \sum_{k=1}^n \frac{1}{(1+\delta)^k}f(x_1,\ldots,x_{k-1},-\frac{x_k}{\delta},0,\ldots,0)$$ and similarly $$f(x)-f(0)\geq -\delta (1+\delta)^n \sum_{k=1}^n \frac{a}{(1+\delta)^k}\geq -an2^n\geq -\epsilon$$