Question

Let \(f_{n} \rightarrow f\) (uniformly) on \(B\). Prove the equivalence of the following statements: (i) Each \(f_{n}\), from a certain \(n\) onward, is bounded on \(B\). (ii) \(f\) is bounded on \(B\). (iii) The \(f_{n}\) are ultimately uniformly bounded on \(B ;\) that is, all function values \(f_{n}(x), x \in B,\) from a certain \(n=n_{0}\) onward, are in one and the same globe \(G_{q}(K)\) in the range space. For real, complex, and vector-valued functions, this means that $$ \left(\exists K \in E^{1}\right)\left(\forall n \geq n_{0}\right)(\forall x \in B) \quad\left|f_{n}(x)\right|

Short answer

Statements (i), (ii), and (iii) are equivalent under uniform convergence.

Step-by-step solution

Understanding the Uniform Convergence

We start with the definition of uniform convergence. A sequence of functions \( f_n \) converges uniformly to a function \( f \) on a set \( B \) if for every \( \epsilon > 0 \), there exists an integer \( N \) such that for all \( x \in B \) and for all \( n \geq N \), \( |f_n(x) - f(x)| < \epsilon \). This definition will be useful in proving the equivalence of the statements.

Prove (i) implies (ii)

Assuming (i), which means each \( f_n \), from some \( n \) onward, is bounded on \( B \). Thus, there exists \( M > 0 \) such that \( |f_n(x)| < M \) for all \( n \geq n_0 \) and all \( x \in B \). Since \( f_n \rightarrow f \) uniformly, there exists an \( N\) such that for all \( n \geq N \), \(|f_n(x) - f(x)| < 1\). It follows that \(|f(x)| \leq |f_n(x)| + |f_n(x) - f(x)| < M + 1\), so \( f \) is bounded.

Prove (ii) implies (iii)

Assuming (ii), we know \( f \) is bounded on \( B \). Therefore, there exists a constant \( K > 0 \) such that \( |f(x)| < K \) for all \( x \in B \). Since \( f_n \rightarrow f \) uniformly, there exists an \( N \) such that for all \( x \in B \) and all \( n \geq N \), \( |f_n(x) - f(x)| < 1 \). Hence, \( |f_n(x)| < |f(x)| + 1 < K + 1 \) for all \( x \in B \) and \( n \geq N \), proving (iii).

Prove (iii) implies (i)

Assuming (iii), the functions \( f_n \) are ultimately uniformly bounded on \( B \), meaning \( |f_n(x)| < K \) for all \( n \geq n_0 \) and all \( x \in B \). Therefore, starting from \( n_0 \), each \( f_n \) is bounded on \( B \) by the constant \( K \), so (i) holds.

Key concepts

Bounded Function

A bounded function is simply a function that doesn’t go to infinity. For any function defined on a specific set, if it doesn't exceed certain bounds, we call it "bounded". Think of having a rubber band stretched between two fingers. No matter how stretchy it gets, it stays within those two points.
A function is bounded if:

• There exists a real number, say, \( M \), such that the absolute value of the function is always less than \( M \) for every input in the domain.
• Mathematically, for a function \( f(x) \) defined on some set \( B \), \( f(x) \) is bounded if \(|f(x)| < M\), where \( M \) is a real number.

In the context of this problem, if each piece in a sequence of functions is ultimately bounded, so is the limiting function. It simply means if all the functions in the sequence don't "blow up" to infinity, the one they converge to won't either. Simple as that!

Sequence of Functions

A sequence of functions is like lining up functions in a row, each waiting its turn. Imagine you have a sequence of numbers like 1, 2, 3, and so on. Now, think about each number at a position being replaced by a function.
Here's how it relates to functions:

• Instead of numbers forming a sequence, you have functions \( f_1, f_2, f_3, \ldots \).
• This sequence can converge, similar to how a sequence of numbers can approach a limit.

In this exercise, we consider uniform convergence, where all functions in the sequence think of reaching a target function \( f \) with minimal margin for error, from a certain point onward. It's like all the runners making sure to finish together, hand in hand, no one straying too far from the leader function.

Real and Complex Functions

Real and complex functions involve either real numbers or complex numbers in their outputs. - Real functions deal purely with real numbers—numbers you number line.- Complex functions, on the other hand, include numbers like a combination of a real and an imaginary part, often expressed as \( a + bi \), where \( i \) is the square root of -1.
When it comes back to our context:

• For real functions, any bounding or limiting is purely about staying within real numbers.
• For complex functions, we are making sure both real and imaginary parts stay within bounds.

The uniform convergence in these functions needs to address both the real part and the imagine part—like making sure your spaghetti is perfectly cooked, nose and tail!

Vector-valued Functions

Vector-valued functions take parameters like a regular function but return vectors. Imagine it like arrows pointing in 3D space, each function calling one set direction (like North, East, or Up). These functions map from a space to vectors, often used to describe physical phenomena like velocity or force.
The key things about vector-valued functions:

• Their outputs are vectors which have both magnitude and direction.
• They can be multi-dimensional (more dimensions than just one or two).

In the problem, these functions must ultimately be uniformly bounded, meaning their sizes (determined by vector length) don’t go beyond a certain boundary. It's like making sure all the arrows in a quiver don't extend beyond a certain length so they all can fit into the target zone without one sticking out awkwardly!