Function Analysis on the Domain [0,1]\(^2\)

Let the function \(f: [0,1]^2 \to \mathbb{R}\) be defined by:

\[ f(x, y) = \begin{cases} \frac{xy}{x^2+y^2} & \text{if } (x, y) \neq (0, 0) \\ 0 & \text{if } x = y = 0 \end{cases} \]

We need to determine the truthfulness of several statements regarding this function on the domain \([0,1]^2\).

Continuity at (0, 0)

To check for continuity at \((0,0)\), we need to see if the limit of \(f(x,y)\) as \((x,y) \to (0,0)\) exists and is equal to \(f(0,0)\). We check the limit along different paths approaching \((0,0)\).

Consider approaching along the line \(y = mx\):

\[ \lim_{(x,y) \to (0,0) \text{ along } y=mx} f(x,y) = \lim_{x \to 0} f(x, mx) = \lim_{x \to 0} \frac{x(mx)}{x^2 + (mx)^2} = \lim_{x \to 0} \frac{mx^2}{x^2(1 + m^2)} \]

For \(x \neq 0\), we can cancel \(x^2\):

\[ \lim_{x \to 0} \frac{m}{1 + m^2} = \frac{m}{1 + m^2} \]

The value of the limit depends on the slope \(m\) of the line of approach. For example, along the x-axis (\(m=0\)), the limit is \(0\), but along the line \(y=x\) (\(m=1\)), the limit is \(\frac{1}{1+1} = \frac{1}{2}\).

Since the limit depends on the path, the limit \(\lim_{(x,y) \to (0,0)} f(x,y)\) does not exist. For continuity at \((0,0)\), the limit must exist and equal \(f(0,0) = 0\). Therefore, \(f\) is not continuous at \((0,0)\).

Statement 1: \(f\) is continuous at \((0, 0)\) - This statement is false.

Bounded Function

A function is bounded on a domain if there exists a real number \(M\) such that \(|f(x,y)| \le M\) for all \((x,y)\) in the domain. The domain here is the closed rectangle \([0,1]^2\).

For \((x,y) \neq (0,0)\), we have \(f(x,y) = \frac{xy}{x^2+y^2}\).

We can use the inequality \(2|xy| \le x^2 + y^2\) (which follows from \((|x|-|y|)^2 \ge 0 \Rightarrow x^2 - 2|xy| + y^2 \ge 0 \Rightarrow x^2+y^2 \ge 2|xy|\)).

For \((x,y) \neq (0,0)\), \(x^2+y^2 > 0\). Dividing by \(x^2+y^2\), we get:

\[ \frac{2|xy|}{x^2+y^2} \le 1 \implies \left|\frac{xy}{x^2+y^2}\right| \le \frac{1}{2} \]

So, for all \((x,y) \neq (0,0)\), \(|f(x,y)| \le \frac{1}{2}\).

At \((0,0)\), \(f(0,0) = 0\), and \(|f(0,0)| = 0 \le \frac{1}{2}\).

Thus, for all \((x,y) \in [0,1]^2\), \(|f(x,y)| \le \frac{1}{2}\). The function \(f\) is bounded on \([0,1]^2\) by \(M = \frac{1}{2}\).

Statement 2: \(f\) is a bounded function - This statement is true.

Existence of Double Integral

The question asks if the double integral \(\int^1_0\int^1_0 f(x,y)dxdy\) exists.

For a function defined on a closed rectangle (like \([0,1]^2\)), the Riemann integral exists if the function is bounded on the rectangle and its set of discontinuities has Lebesgue measure zero.

We have established that \(f\) is bounded on \([0,1]^2\).

The function \(f(x,y) = \frac{xy}{x^2+y^2}\) is a rational function for \((x,y) \neq (0,0)\). It is continuous at all points where the denominator is non-zero. The denominator \(x^2+y^2\) is zero only at \((0,0)\). So, \(f\) is continuous on \([0,1]^2 \setminus \{(0,0)\}\).

The only point where continuity might fail is \((0,0)\). We already showed it is not continuous at \((0,0)\).

The set of discontinuities of \(f\) on \([0,1]^2\) is the single point \(\{(0,0)\}\). A single point in \(\mathbb{R}^2\) has Lebesgue measure zero.

Since \(f\) is bounded on \([0,1]^2\) and its set of discontinuities has measure zero, the double integral \(\int^1_0\int^1_0 f(x,y)dxdy\) exists.

Statement 3: \(\int^1_0\int^1_0 f(x,y)dxdy\) exists - This statement is true.

Continuity at (1, 0)

To check for continuity at \((1,0)\), we need to see if the limit of \(f(x,y)\) as \((x,y) \to (1,0)\) exists and is equal to \(f(1,0)\).

The point \((1,0)\) is in the domain \([0,1]^2\). Since \((1,0) \neq (0,0)\), the function is defined by \(f(x,y) = \frac{xy}{x^2+y^2}\) for points near \((1,0)\).

The limit as \((x,y) \to (1,0)\) of \(f(x,y)\) is:

\[ \lim_{(x,y) \to (1,0)} \frac{xy}{x^2+y^2} \]

Since the denominator \(x^2+y^2\) is non-zero at \((1,0)\) (\(1^2+0^2=1\)), and the expression is a ratio of polynomials, the limit can be found by direct substitution:

\[ \lim_{(x,y) \to (1,0)} \frac{xy}{x^2+y^2} = \frac{(1)(0)}{1^2+0^2} = \frac{0}{1} = 0 \]

Now we find the function value at \((1,0)\). Since \((1,0) \neq (0,0)\), we use the first part of the definition:

\[ f(1,0) = \frac{(1)(0)}{1^2+0^2} = \frac{0}{1} = 0 \]

Since \(\lim_{(x,y) \to (1,0)} f(x,y) = 0\) and \(f(1,0) = 0\), the limit exists and is equal to the function value. Therefore, \(f\) is continuous at \((1,0)\).

Statement 4: \(f\) is continuous at \((1, 0)\) - This statement is true.

Summary of True Statements

Based on our analysis, the following statements are true:

• \(f\) is a bounded function.
• \(\int^1_0\int^1_0 f(x,y)dxdy\) exists.
• \(f\) is continuous at \((1, 0)\).

These correspond to statements 2, 3, and 4 from the options.