A function $f \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ iscoordinate-wise convex if $f_x \colon \mathbb{R} \rightarrow> \mathbb{R}, t \rightarrow f(x, t)$ and $f_y \colon \mathbb{R}> \rightarrow \mathbb{R}, t \rightarrow f(t, y)$ are convex for any $x,> y$ in $\mathbb{R}$. Prove that $$\underline{\lim}\limits_{t> \rightarrow 0+} \frac{f(t, t) - f(0, 0)}{t} +> \underline{\lim}\limits_{t \rightarrow 0-} \frac{f(0, 0) - f(t, t)}{t}> \geq 0$$
I think this can be done by considering some $x, y$ and showing that$$\underline{\lim}\limits_{t \rightarrow 0+} \frac{f(x, t) - f(0, 0)}{t} + \underline{\lim}\limits_{t \rightarrow 0-} \frac{f(0, 0) - f(t, y)}{t} \geq 0$$Then I can take a limit over $x, y$ and get the original claim. But I don't know how to write this down correctly (and if the idea is even correct). Thanks!