Understanding the Problem

We are given a function \(f:\mathbb{R}^2\rightarrow\mathbb{R}\) that is bounded. This means there exist real numbers \(M\) and \(N\) such that \(N \le f(x,y) \le M\) for all \((x,y) \in \mathbb{R}^2\).

Additionally, the function has specific monotonicity properties:

• For each fixed \(t \in \mathbb{R}\), the function \(g_t(y) = f(t, y)\) is non-decreasing with respect to \(y\). This means if \(y_1 \le y_2\), then \(f(t, y_1) \le f(t, y_2)\).
• For each fixed \(t \in \mathbb{R}\), the function \(h_t(x) = f(x, t)\) is non-decreasing with respect to \(x\). This means if \(x_1 \le x_2\), then \(f(x_1, t) \le f(x_2, t)\).

We need to evaluate which of the given statements are necessarily true based on these properties.

Analyzing Option 1: Monotonicity of k(x) = f(x, x)

Let \(k(x) = f(x, x)\). We want to check if \(k(x)\) is a non-decreasing function. This means we need to check if for any \(x_1, x_2 \in \mathbb{R}\) with \(x_1 \le x_2\), we have \(k(x_1) \le k(x_2)\), which is \(f(x_1, x_1) \le f(x_2, x_2)\).

Let's use the given properties. Since \(x_1 \le x_2\), and for a fixed \(y\)-value (let's use \(y=x_1\)), the function \(h_{x_1}(x) = f(x, x_1)\) is non-decreasing in \(x\), we have:

\(f(x_1, x_1) \le f(x_2, x_1)\)

Now, consider the \(x\)-value \(x_2\). Since \(x_1 \le x_2\), and for a fixed \(x\)-value (let's use \(x=x_2\)), the function \(g_{x_2}(y) = f(x_2, y)\) is non-decreasing in \(y\), we have:

\(f(x_2, x_1) \le f(x_2, x_2)\)

Combining these two inequalities, we get:

\(f(x_1, x_1) \le f(x_2, x_1) \le f(x_2, x_2)\)

Thus, if \(x_1 \le x_2\), then \(f(x_1, x_1) \le f(x_2, x_2)\). This shows that the function \(k(x) = f(x, x)\) is indeed a non-decreasing function.

Statement 1 is necessarily true.

Analyzing Option 3: Existence of Limit at (\(\boldsymbol{+\infty, +\infty}\))

We are asked if \({\lim _{\left( {x,y} \right) \to \left( { + \infty , + \infty } \right)}}f\left( {x,y} \right)\) exists.

Let \((x_1, y_1)\) and \((x_2, y_2)\) be two points such that \(x_1 \le x_2\) and \(y_1 \le y_2\). Using the property that \(f(x, t)\) is non-decreasing in \(x\) for fixed \(t\), we have \(f(x_1, y_1) \le f(x_2, y_1)\) since \(x_1 \le x_2\). Using the property that \(f(t, y)\) is non-decreasing in \(y\) for fixed \(t\), we have \(f(x_2, y_1) \le f(x_2, y_2)\) since \(y_1 \le y_2\).

Combining these, we get \(f(x_1, y_1) \le f(x_2, y_1) \le f(x_2, y_2)\). This shows that \(f\) is non-decreasing with respect to the product order on \(\mathbb{R}^2\), where \((x_1, y_1) \le (x_2, y_2)\) if and only if \(x_1 \le x_2\) and \(y_1 \le y_2\).

We are considering the limit as \((x,y) \to (+\infty, +\infty)\), which means \(x \to +\infty\) and \(y \to +\infty\). As \(x\) and \(y\) increase towards infinity, the value of \(f(x,y)\) is non-decreasing because for any \((x_1, y_1)\) and \((x_2, y_2)\) with \(x_1 \le x_2\) and \(y_1 \le y_2\), we have \(f(x_1, y_1) \le f(x_2, y_2)\). The function \(f\) is also bounded above by \(M\).

A non-decreasing function (with respect to the product order) that is bounded above on \(\mathbb{R}^2\) must converge to a limit as \((x,y) \to (+\infty, +\infty)\). The limit is equal to the supremum of the function values.

Statement 3 is necessarily true.

Analyzing Option 2: Number of Discontinuities

The statement claims the number of discontinuities of \(f\) is at most countably infinite. While a non-decreasing function of one variable has at most countably many discontinuities, this property does not directly extend to functions of two variables that are only non-decreasing in each variable separately. A function can be non-decreasing in each variable separately but still have an uncountable set of discontinuities.

Consider the function \(f(x,y) = 0\) if \(x \le 0\) or \(y \le 0\), and \(f(x,y) = 1\) if \(x > 0\) and \(y > 0\). This function is bounded (between 0 and 1).

• Fix \(t\). \(g_t(y) = f(t,y)\). If \(t \le 0\), \(g_t(y) = 0\) for all \(y\), which is non-decreasing. If \(t > 0\), \(g_t(y) = 0\) for \(y \le 0\) and \(g_t(y) = 1\) for \(y > 0\), which is non-decreasing in \(y\).
• Fix \(t\). \(h_t(x) = f(x,t)\). If \(t \le 0\), \(h_t(x) = 0\) for all \(x\), which is non-decreasing. If \(t > 0\), \(h_t(x) = 0\) for \(x \le 0\) and \(h_t(x) = 1\) for \(x > 0\), which is non-decreasing in \(x\).

This function satisfies all the given conditions. However, it is discontinuous at every point on the positive x-axis \(\{(x, 0) : x > 0\}\) and every point on the positive y-axis \(\{(0, y) : y > 0\}\). The set of discontinuities contains segments of lines, which is uncountable. Therefore, the number of discontinuities is not necessarily at most countably infinite.

Statement 2 is not necessarily true.

Analyzing Option 4: Existence of Limit at (\(\boldsymbol{+\infty, -\infty}\))

We are asked if \({\lim _{\left( {x,y} \right) \to \left( { + \infty , - \infty } \right)}}f\left( {x,y} \right)\) exists.

Let \(x \to +\infty\) and \(y \to -\infty\). This means \(x\) is increasing while \(y\) is decreasing. The monotonicity property with respect to the product order (\(f(x_1, y_1) \le f(x_2, y_2)\) if \(x_1 \le x_2\) and \(y_1 \le y_2\)) does not apply directly here because \(y\) is decreasing.

Consider the function \(f(x,y) = \arctan(x+y)\). This function is bounded (between \(-\pi/2\) and \(\pi/2\)). We checked earlier that it is non-decreasing in \(x\) for fixed \(y\) and non-decreasing in \(y\) for fixed \(x\).

Let's evaluate the limit as \((x,y) \to (+\infty, -\infty)\) along different paths:

• Path 1: Let \(y = -x\). As \(x \to +\infty\), \(y \to -\infty\). \(x+y = x+(-x) = 0\). \(f(x,y) = \arctan(0) = 0\). The limit along this path is 0.
• Path 2: Let \(y = -x+1\). As \(x \to +\infty\), \(y \to -\infty\). \(x+y = x+(-x+1) = 1\). \(f(x,y) = \arctan(1) = \pi/4\). The limit along this path is \(\pi/4\).

Since the limit depends on the path taken to \((+\infty, -\infty)\), the limit \({\lim _{\left( {x,y} \right) \to \left( { + \infty , - \infty } \right)}}f\left( {x,y} \right)\) does not exist for this function. Since this function satisfies the given conditions, Statement 4 is not necessarily true.

Conclusion

Based on our analysis:

• Statement 1 is necessarily true.
• Statement 2 is not necessarily true.
• Statement 3 is necessarily true.
• Statement 4 is not necessarily true.

The statements that are necessarily true are 1 and 3.