Understanding the Functions f and g

The problem defines two functions, f and g. The function \(f: \mathbb{R}^2 \to \mathbb{R}\) is given by:

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

This function is well-known in multivariable calculus. It is continuous everywhere except possibly at the origin (0,0). At (0,0), the limit of f depends on the path of approach. For example, approaching along the line y = mx, we get \(\lim_{(x,mx) \to (0,0)} \frac{2x(mx)}{x^2+(mx)^2} = \lim_{x \to 0} \frac{2mx^2}{x^2(1+m^2)} = \frac{2m}{1+m^2}\). Since this limit depends on m, the limit as (x,y) \(\to\) (0,0) does not exist, and thus f is discontinuous at (0,0).

The function g is defined as an infinite series:

\[ g(x, y) = \sum_{n=1}^{\infty} \frac{f((x-n),(y-n))}{2^n} \]

Let's analyze the properties of g based on this definition.

Analyzing the Convergence of g(x,y)

Consider the terms of the series defining g(x,y). Each term is \(\frac{f((x-n),(y-n))}{2^n}\). We know that for \((u,v) \neq (0,0)\), \(|f(u,v)| = \left|\frac{2uv}{u^2+v^2}\right|\). Since \((u-v)^2 = u^2 - 2uv + v^2 \ge 0\), we have \(2uv \le u^2 + v^2\). Also, \((u+v)^2 = u^2 + 2uv + v^2 \ge 0\), so \(-2uv \le u^2 + v^2\). Combining these, \(|2uv| \le u^2 + v^2\). Thus, for \((u,v) \neq (0,0)\), \(|f(u,v)| = \frac{|2uv|}{u^2+v^2} \le \frac{u^2+v^2}{u^2+v^2} = 1\). At (0,0), \(f(0,0)=0\), so \(|f(0,0)|=0 \le 1\). Therefore, \(|f(u,v)| \le 1\) for all \((u,v) \in \mathbb{R}^2\).

The terms of the series for g(x,y) satisfy:

\[ \left|\frac{f((x-n),(y-n))}{2^n}\right| \le \frac{1}{2^n} \]

The series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) is a geometric series with ratio 1/2, which converges to 1. By the Weierstrass M-test, the series \(\sum_{n=1}^{\infty} \frac{f((x-n),(y-n))}{2^n}\) converges uniformly for all \((x,y) \in \mathbb{R}^2\). Uniform convergence implies that g(x,y) is a well-defined function on \(\mathbb{R}^2\).

Evaluating Option 3: g is not a well-defined function

Since the series defining g(x,y) converges uniformly on \(\mathbb{R}^2\), the function g is well-defined for all \((x,y) \in \mathbb{R}^2\). Therefore, Option 3 is false.

Analyzing Continuity of g(x,y)

The uniform convergence of the series implies that g is continuous wherever each term \(T_n(x,y) = \frac{f((x-n),(y-n))}{2^n}\) is continuous. The term \(T_n(x,y)\) is continuous unless the argument of f, \((x-n, y-n)\), is equal to (0,0). This happens when \(x-n = 0\) and \(y-n = 0\), i.e., when \(x=n\) and \(y=n\). So, the n-th term \(T_n(x,y)\) is discontinuous only at the point (n, n).

Evaluating Option 4: g is continuous on \(\mathbb{R}^2 \setminus \{(k, k)\}_{k \in \mathbb{N}}\)

Let's consider a point \((a,b) \in \mathbb{R}^2\). If \((a,b) \neq (k,k)\) for any positive integer k, then for every \(n \in \mathbb{N}\), \((a-n, b-n) \neq (0,0)\). Since f is continuous everywhere except at (0,0), the function \(f((x-n),(y-n))\) is continuous in a neighborhood around \((a,b)\) for each n. Since the series \(\sum_{n=1}^{\infty} \frac{f((x-n),(y-n))}{2^n}\) converges uniformly and each term is continuous in a neighborhood of \((a,b)\), the sum g(x,y) is continuous at \((a,b)\).

Now, let's consider a point \((k,k)\) where \(k \in \mathbb{N}\). The k-th term, \(T_k(x,y) = \frac{f((x-k),(y-k))}{2^k}\), is discontinuous at \((k,k)\) because f is discontinuous at (0,0).

Let's calculate \(g(k,k)\):

\[ g(k, k) = \sum_{n=1}^{\infty} \frac{f((k-n),(k-n))}{2^n} \]

• For \(n=k\), \((k-n, k-n) = (0,0)\), so \(f((k-k),(k-k)) = f(0,0) = 0\).
• For \(n \neq k\), \((k-n, k-n) \neq (0,0)\). Let \(u=k-n, v=k-n\). Then \(f((k-n),(k-n)) = f(u,v) = \frac{2uv}{u^2+v^2} = \frac{2(k-n)^2}{(k-n)^2+(k-n)^2} = \frac{2(k-n)^2}{2(k-n)^2} = 1\).

So,

\[ g(k,k) = \sum_{n=1, n \neq k}^{\infty} \frac{1}{2^n} + \frac{f(0,0)}{2^k} = \sum_{n=1, n \neq k}^{\infty} \frac{1}{2^n} + 0 \]

The sum of the full series is \(\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1/2}{1-1/2} = 1\). Thus, \(\sum_{n=1, n \neq k}^{\infty} \frac{1}{2^n} = \sum_{n=1}^{\infty} \frac{1}{2^n} - \frac{1}{2^k} = 1 - \frac{1}{2^k}\). So, \(g(k,k) = 1 - \frac{1}{2^k}\).

Now consider the limit of g(x,y) as \((x,y) \to (k,k)\). Let's approach along the line y=x. Let \(x=k+u\) and \(y=k+u\). As \(u \to 0\), \((x,y) \to (k,k)\) along this path.

\[ g(k+u, k+u) = \sum_{n=1}^{\infty} \frac{f((k+u-n),(k+u-n))}{2^n} \]

• For \(n=k\), the term is \(\frac{f(u,u)}{2^k}\). For \(u \neq 0\), \(f(u,u) = \frac{2u^2}{u^2+u^2} = 1\). So \(\lim_{u \to 0, u \neq 0} \frac{f(u,u)}{2^k} = \frac{1}{2^k}\).
• For \(n \neq k\), let \(m=k-n \neq 0\). The term is \(\frac{f(m+u, m+u)}{2^n}\). As \(u \to 0\), \((m+u, m+u) \to (m,m)\). Since \(m \neq 0\), \(f\) is continuous at \((m,m)\). So \(\lim_{u \to 0} \frac{f(m+u, m+u)}{2^n} = \frac{f(m,m)}{2^n} = \frac{1}{2^n}\).

By uniform convergence, we can take the limit inside the sum for terms where the limit exists and is continuous. The difficulty is the k-th term. Let's split the sum:

\[ \lim_{(x,y) \to (k,k), y=x} g(x,y) = \lim_{u \to 0} \left( \frac{f(u,u)}{2^k} + \sum_{n=1, n \neq k}^{\infty} \frac{f((k+u-n),(k+u-n))}{2^n} \right) \]

\[ = \lim_{u \to 0, u \neq 0} \frac{f(u,u)}{2^k} + \sum_{n=1, n \neq k}^{\infty} \lim_{u \to 0} \frac{f((k+u-n),(k+u-n))}{2^n} \] (by uniform convergence of the remaining sum)

\[ = \frac{1}{2^k} + \sum_{n=1, n \neq k}^{\infty} \frac{f((k-n),(k-n))}{2^n} \] \[ = \frac{1}{2^k} + \sum_{n=1, n \neq k}^{\infty} \frac{1}{2^n} = \frac{1}{2^k} + \left(\sum_{n=1}^{\infty} \frac{1}{2^n} - \frac{1}{2^k}\right) = \frac{1}{2^k} + (1 - \frac{1}{2^k}) = 1 \]

Since \(\lim_{(x,y) \to (k,k), y=x} g(x,y) = 1\) and \(g(k,k) = 1 - \frac{1}{2^k}\), and \(1 \neq 1 - \frac{1}{2^k}\) for \(k \in \mathbb{N}\), the function g is discontinuous at each point \((k,k)\) for \(k \in \mathbb{N}\). Combining this with the continuity at points not of the form (k,k), we conclude that g is continuous on \(\mathbb{R}^2 \setminus \{(k, k)\}_{k \in \mathbb{N}}\). Therefore, Option 4 is true.

Evaluating Option 2: g is continuous from \(\mathbb{R}^2\) into \(\mathbb{R}\)

Since g is discontinuous at the points \((k, k)\) for \(k \in \mathbb{N}\), it is not continuous on all of \(\mathbb{R}^2\). Therefore, Option 2 is false.

Evaluating Option 1: The function h(y) = g(c, y) is continuous on \(\mathbb{R}\) for all c

Let's fix a value c and consider the function \(h(y) = g(c, y) = \sum_{n=1}^{\infty} \frac{f((c-n),(y-n))}{2^n}\). The terms of this series are \(S_n(y) = \frac{f((c-n),(y-n))}{2^n}\). For this term to be discontinuous, the argument of f, \((c-n, y-n)\), must be (0,0). This means \(c-n=0\) and \(y-n=0\), i.e., \(c=n\) and \(y=n\).

• Case 1: c is not a positive integer. In this case, \(c-n\) is never zero for any \(n \in \mathbb{N}\). So, for any y, \((c-n, y-n)\) is never (0,0). Thus, \(f((c-n),(y-n))\) is continuous as a function of y for each n. Since the series converges uniformly, the sum \(h(y) = g(c, y)\) is continuous for all y.
• Case 2: c is a positive integer, say \(c=k\) for some \(k \in \mathbb{N}\). The series is \(h(y) = g(k, y) = \sum_{n=1}^{\infty} \frac{f((k-n),(y-n))}{2^n}\). Consider the k-th term: \(S_k(y) = \frac{f((k-k),(y-k))}{2^k} = \frac{f(0, y-k)}{2^k}\). We know \(f(0,v) = 0\) for all v. So \(f(0, y-k) = 0\) for all y. Thus, the k-th term \(S_k(y) = \frac{0}{2^k} = 0\) for all y. This term is continuous. Now consider the terms for \(n \neq k\). The term is \(S_n(y) = \frac{f((k-n),(y-n))}{2^n}\). For this term to be discontinuous, \((k-n, y-n)\) must be (0,0), which implies \(k=n\) and \(y=n\). Since we are considering \(n \neq k\), the first condition \(k=n\) is never met. Thus, for \(n \neq k\), the point \((k-n, y-n)\) is never (0,0) for any y. So, \(f((k-n),(y-n))\) is continuous as a function of y for each \(n \neq k\).

So, when c=k (a positive integer), \(h(y) = g(k, y) = \sum_{n=1, n \neq k}^{\infty} \frac{f((k-n),(y-n))}{2^n} + S_k(y) = \sum_{n=1, n \neq k}^{\infty} \frac{f((k-n),(y-n))}{2^n} + 0\). The series \(\sum_{n=1, n \neq k}^{\infty} \frac{f((k-n),(y-n))}{2^n}\) is a uniformly convergent series of continuous functions of y. Therefore, its sum is continuous for all y.

In both cases (c not a positive integer and c a positive integer), h(y) = g(c, y) is continuous on \(\mathbb{R}\) for all y. Therefore, Option 1 is true.

Conclusion

Based on our analysis:

• Option 1: The function h(y) = g(c, y) is continuous on \(\mathbb{R}\) for all c - True.
• Option 2: g is continuous from \(\mathbb{R}^2\) into \(\mathbb{R}\) - False.
• Option 3: g is not a well-defined function - False.
• Option 4: g is continuous on \(\mathbb{R}^2 \setminus \{(k, k)\}_{k \in \mathbb{N}}\) - True.

The statements that are true are Option 1 and Option 4.