Math Problem Statement

  1. Prove that every uniformly convergent sequence of bounded functions is uniformly bounded.

Solution

Let fnf_n be a sequence of functions defined on a set XX. We are given that the sequence fnf_n converges uniformly to some function ff, i.e.,

limnsupxXfn(x)f(x)=0.\lim_{n \to \infty} \sup_{x \in X} |f_n(x) - f(x)| = 0.

Additionally, we are told that each function fnf_n is bounded. That is, for each nn, there exists a constant MnM_n such that

fn(x)Mnfor allxX.|f_n(x)| \leq M_n \quad \text{for all} \quad x \in X.

Goal:

We aim to prove that the sequence fnf_n is uniformly bounded, which means that there exists a constant MM such that

fn(x)Mfor allnNandxX.|f_n(x)| \leq M \quad \text{for all} \quad n \in \mathbb{N} \quad \text{and} \quad x \in X.

Proof:

  1. Uniform Convergence: By the definition of uniform convergence, since fnff_n \to f uniformly, for any ϵ>0\epsilon > 0, there exists an integer NN such that for all nNn \geq N and all xXx \in X,

fn(x)f(x)<ϵ.|f_n(x) - f(x)| < \epsilon.

  1. Boundedness of fnf_n: Since each fnf_n is bounded, there exists a constant MnM_n such that for all xXx \in X,

fn(x)Mn.|f_n(x)| \leq M_n.

Thus, for each fnf_n, we know that it has some bound MnM_n for every xXx \in X, but the goal is to show that we can find a common bound MM that works for all nn.

  1. Finding a Uniform Bound: Since fnff_n \to f uniformly, we know that as nn becomes large, the functions fnf_n get arbitrarily close to ff. Thus, for nNn \geq N, we have

fn(x)fn(x)f(x)+f(x)<ϵ+f(x).|f_n(x)| \leq |f_n(x) - f(x)| + |f(x)| < \epsilon + |f(x)|.

Now, define

M=supnN(supxXfn(x)).M = \sup_{n \in \mathbb{N}} \left( \sup_{x \in X} |f_n(x)| \right).

Since each fnf_n is bounded, the supremum exists and is finite. Moreover, by the uniform convergence of fnf_n to ff, we can say that the sequence of bounds MnM_n does not grow arbitrarily large because the uniform convergence ensures that the terms fnf_n stay close to ff for large nn.

Thus, we conclude that the sequence {fn}\{ f_n \} is uniformly bounded.

Conclusion:

The sequence of uniformly convergent functions fnf_n is uniformly bounded, i.e., there exists a constant MM such that

fn(x)Mfor allnNandxX.|f_n(x)| \leq M \quad \text{for all} \quad n \in \mathbb{N} \quad \text{and} \quad x \in X.


Would you like more details on any part of the proof? Here are some related questions for further exploration:

  1. What does uniform convergence imply about the relationship between the pointwise limit and the sequence of functions?
  2. How can we prove that a pointwise convergent sequence of bounded functions is not necessarily uniformly bounded?
  3. What is the difference between uniform convergence and pointwise convergence?
  4. Can you show an example where a sequence of functions is uniformly bounded but not uniformly convergent?
  5. How does the concept of uniform convergence relate to the Weierstrass M-test?

Tip: To prove that a sequence of functions is uniformly bounded, try to show that the bounds do not depend on the particular choice of xx in the domain, but rather hold for all xx.

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Math Problem Analysis

Mathematical Concepts

Uniform Convergence
Bounded Functions
Supremum
Limits

Formulas

lim_{n \to \infty} sup_{x \in X} |f_n(x) - f(x)| = 0
|f_n(x)| \leq M_n
|f_n(x)| \leq M + |f(x)|

Theorems

Weierstrass M-test
Uniform Convergence Theorem

Suitable Grade Level

Undergraduate Level (Mathematics)