Consider the following two-variable function:
\begin{equation}
f(x,t)=h(x)+g(t)
\end{equation}
assume the following relation holds for $0\leq\lambda\leq 0$:
\begin{align}
&h(\lambda x +(1-\lambda) y)\leq \lambda h(x)+(1-\lambda)h(y)\notag\\
&g(\lambda t +(1-\lambda) s)\leq \lambda g(t)+(1-\lambda)g(s)
\end{align}
can we conclude that the function $f(x,t)$ is also convex?
more generally if we have an arbitrary two-variable function $k(x,t)$ which is convex w.r.t its individual arguments, can we conclude that it is convex as a multivariable function as well? i.e.
if
\begin{align}
&k\Big((\lambda x +(1-\lambda) y),t\Big)\leq \lambda k(x,t)+(1-\lambda)k(y,t)\notag\\
&k\Big(x,(\lambda t +(1-\lambda) s)\Big)\leq \lambda k(x,t)+(1-\lambda)k(x,s)
\end{align}
Can we conclude that
\begin{align}
&k\Big((\lambda x +(1-\lambda) y),(\lambda t +(1-\lambda) s)\Big)\leq \lambda k(x,t)+(1-\lambda)k(y,s)
\end{align}
Best Answer
The answer to the first question is “yes”: Both functions $$ (x, t) \mapsto h(x) \\ (x, t) \mapsto g(t) $$ are convex, and the sum of convex functions is convex.
The answer to the second question is “no”: A counterexample is $f(x, t) = xt$, which is linear (and therefore convex) in both variables separately, but not convex: $$ 1 = f(1, 1) \le \frac 12 \left( f(2, 0) + f(0, 2) \right) = 0 $$ does not hold.