Conditions for $f(x_n,y_n) - f(x_0, y_n) \to 0$ if $(x_n,y_n) \to (x_0,y_0)$?

Question

I have $x_n \to x_0$ and $y_n \to y_0$ as $n \to \infty$. Let $f(x,y)$ be a real-valued function.

Question: Under which conditions do I have

$$f(x_n, y_n) - f(x_0, y_n) \to 0 $$

as $n \to \infty$?

I know two separate conditions, but I was wondering if there were other less restrictive assumptions for the convergence listed above.

Condition 1: There is a neighborhood $\mathcal{B}$ of $y_0$ on which $f(x,y)$ is continuous in $x$ uniformly in $y$, i.e

$$ \sup_{y \in \mathcal{B}} \left|f(x_n, y) - f(x_0, y)\right| \to 0 $$

in which case we have, for $n$ larger enough so that $y_n \in \mathcal{B}$,

$$|f(x_n, y_n) - f(x_0, y_n)| \le \sup_{y \in \mathcal{B}} \left|f(x_n, y) - f(x_0, y)\right| \to 0$$

Alternatively,

Condition 2: $f(x,y)$ is continuous at $(x_0,y_0)$.

We then have, through the triangle inequality,

$$\begin{align}|f(x_n, y_n) - f(x_0, y_n)| &\le |f(x_n, y_n) - f(x_0, y_0)| + |f(x_0, y_n) - f(x_0, y_0)| \to 0 \end{align}$$

Neither condition implies the other. For example, the following condition satisfies Condition 1 but not Condition 2 for $(x_0,y_0) = (0,0)$,

Example 1: $$ f(x,y) = \left\{\begin{align} 1 , & \text{ for } y \in \mathbb{Q}\\ -1 , & \text{ for } y \notin \mathbb{Q} \end{align}\right. $$

while this function satisfies Condition 2 but not Condition 1,

Example 2: $$ f(x,y) = \left\{\begin{align} y, & \text{ for } x \in \mathbb{Q} \\ - y, & \text{ for } x \notin \mathbb{Q} \end{align}\right. $$

Maybe I am missing a more fundamental condition?

Answer 1

This is not amazingly useful, but I've seen in the literature an assumption similar to the following:

Condition 3: For any sequence of balls $\{ B(y_0, \delta_n )\}$ centered at $x_0 $ of radius $\delta_n \to 0$, we have $$ \sup_{y \in B(y_0,\delta_n)} \left|f(x_n, y) - f(x_0, y)\right| \to 0 $$

That Condition 3 is sufficient can be seenby taking $\{ \delta_n \}$ such that $\delta_n > |y_n - y_0|$ for every $n$.

Clearly Condition 1 implies Condition 3.

Condition 2 also implies Condition 3.

Proof.

First, note that, from the triangle inequality,

$$\begin{align}\sup_{y \in B(y_0,\delta_n)} \left|f(x_n, y) - f(x_0, y)\right| &\le \sup_{y \in B(y_0,\delta_n)} \left|f(x_n, y) - f(x_0, y_0)\right| \\ &+\sup_{y \in B(y_0,\delta_n)} \left|f(x_0, y) - f(x_0, y_0)\right| \end{align}$$

From continuity at $(x_0,y_0)$ both terms can be made arbitrarily small by picking a $n$ large enough.

We can then apply Condition 3 to Example 2.

Example 2 continued: We have

$$ \sup_{y \in B(y_0,\delta_n)} \left|f(x_n, y) - f(x_0, y)\right| = \left\{\begin{align} 0, & \text{ if both $x_n$ and $x_0$ are in $\mathbb{Q}$, or both not in $\mathbb{Q}$ } \\ 2\delta_n & \text{ otherwise } \end{align}\right. $$

which clearly goes to zero.