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Chapter 6: Problem 12

Prove that if a sequence of functions \(f_{n}: S \rightarrow \mathbb{R}\) converge uniformly to a bounded function \(f: S \rightarrow \mathbb{R},\) then there exists an \(N\) such that for all \(n \geq N,\) the \(f_{n}\) are bounded.

Short Answer

Expert verified
For all \(n \geq N\), \(|f_n(x)| \leq 1 + M\), proving \(f_n\) is bounded.

Step by step solution

01

Understanding Uniform Convergence

Uniform convergence of a sequence of functions \(f_n\) to a function \(f\) implies that given any \(\epsilon > 0\), there exists an integer \(N\) such that for all \(n \geq N\) and all \(x \in S\), we have \(|f_n(x) - f(x)| < \epsilon\).
02

Given Condition - Bounded Function

The function \(f\) is bounded means there exists an \(M > 0\) such that for all \(x \in S\), \(|f(x)| \leq M\). This bounding condition is crucial as it indicates the limitation in the range of the values of \(f\).
03

Establishing an Upper Bound for \(f_n\)

Because the functions \(f_n\) converge uniformly to \(f\), choose \(\epsilon = 1\). Then, there exists an \(N\) such that for all \(n \geq N\) and all \(x \in S\), \(|f_n(x) - f(x)| < 1\).
04

Use Triangle Inequality for \(f_n\)

By the triangle inequality, \(|f_n(x)| = |f_n(x) - f(x) + f(x)| \leq |f_n(x) - f(x)| + |f(x)|\). For all \(n \geq N\), this gives us \(|f_n(x)| \leq 1 + M\) for all \(x \in S\).
05

Conclude Boundedness of \(f_n\)

Thus, for all \(n \geq N\), \(f_n\) are bounded with the bound \(1 + M\). Hence, there exists \(N\) such that for all \(n \geq N\), \(|f_n(x)| \leq 1 + M\) for all \(x \in S\).
06

Final Conclusion

Since \(f\) is bounded and \(f_n\) converges uniformly to \(f\), we have shown that each \(f_n\) for \(n \geq N\) is also bounded. This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bounded Function
A function is considered bounded if there is a real number that serves as an upper limit to its absolute value over its entire domain. For a function \( f : S \rightarrow \mathbb{R} \), we say it is bounded if there exists a number \( M > 0 \) such that for all \( x \in S \), the absolute value of \( f(x) \) is less than or equal to \( M \).

The concept of a bounded function is crucial when dealing with uniform convergence. It sets a limit on how far the values of the function can deviate from this limit across its domain.
  • Helpful for understanding behavior over large regions.
  • Makes analyzing limits in terms of another bounded function simpler.
  • Useful for establishing convergence limits for sequences of functions.
If a sequence of functions converges uniformly to a bounded function, the limit function retains this bounded property. This makes it easier to control the behavior of the sequence as it converges.
Sequence of Functions
A sequence of functions is a collection of functions \( \{ f_n \} \) indexed by \( n \), usually taking the form of \( f_n : S \rightarrow \mathbb{R} \). Each function in the sequence can potentially vary with \( n \), offering a dynamic view of how functions behave across inputs.

Uniform convergence is an important property associated with sequences of functions. It ensures that the whole sequence \( \{ f_n \} \) converges to a single function \( f \) at the same rate across the domain \( S \).
  • Allows us to analyze how individual functions in the sequence approach the limiting function \( f \).
  • Crucial for ensuring that properties of the functions are retained in the limit.
  • Ensures that convergence behavior is consistent across the entire domain.
This uniform aspect is vital in many areas of mathematics, ensuring that operations applied to the function and limit are valid across their entire range.
Triangle Inequality
The triangle inequality is a fundamental concept in mathematics which states that for any real numbers \( a \) and \( b \), the inequality \( |a + b| \leq |a| + |b| \) holds. In other words, the absolute value of a sum is less than or equal to the sum of the absolute values.

In the context of function sequences and uniform convergence, this principle plays a key role. It helps in establishing the boundedness criteria for the sequence of functions. Consider the equation:\[|f_n(x) - f(x)| < 1\]If we apply the triangle inequality, we expand and control the absolute distance between \( f_n(x) \) and the boundary established by the function \( f(x) \).
  • It asserts that deviations of \( f_n \) from \( f \) are controlled and limited.
  • Ensures that the function \( f_n \) stays within a bounded region once close to \( f \).
  • Provides a straightforward tool to measure convergence and boundedness.
This concept is integral in proving the uniform convergence of function sequences, especially when combined with the boundedness information of the limit function.

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Most popular questions from this chapter

Let I, \(J \subset \mathbb{R}\) be intervals. Let \(F: I \times J \rightarrow \mathbb{R}\) be a continuous function of two variables and suppose \(f: I \rightarrow J\) be a continuous function. Show that \(F(x, f(x))\) is a continuous function on \(I .\)

A joint limit (see above) does not mean the iterated limits even exist. Consider \(x_{n, m}:=\) \(\frac{(-1)^{n+m}}{\min [n, m]}\) a) Show that for no \(n\) does \(\lim _{m \rightarrow \infty} x_{n, m}\) exist, and for no \(m\) does \(\lim _{n \rightarrow \infty} x_{n, m}\) exist. So neither \(\lim _{n \rightarrow \infty} \lim _{n \rightarrow \infty} x_{n, m}\) nor \(\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty} x_{n, m}\) makes any sense at all. b) Show that the joint limit of \(\left\\{x_{n-m}\right\\}\) exists and is \(0 .\)

A common type of equation one encounters are linear first order differential equations, that is equations of the form $$ y^{\prime}+p(x) y=q(x), \quad y\left(x_{0}\right)=y_{0} . $$ Prove Picard's theorem for linear equations. Suppose I is an interval, \(x_{0} \in I,\) and \(p: I \rightarrow \mathbb{R}\) and \(q: I \rightarrow \mathbb{R}\) are continuous. Show that there exists a unique differentiable \(f: I \rightarrow \mathbb{R},\) such that \(y=f(x)\) satisfies the equation and the initial condition. Hint: Assume existence of the exponential function and use the integrating factor formula for existence of \(f\) (prove that it works): $$ f(x):=e^{-\int_{x_{0}}^{x} p(s) d s}\left(\int_{x_{0}}^{x} e^{\int_{x_{0}}^{t} p(s) d s} q(t) d t+y_{0}\right) $$

The Dirichlet function \(f:[0,1] \rightarrow \mathbb{R}\), that is the function such that \(f(x):=1\) if \(x \in \mathbb{Q}\) and \(f(x):=0\) if \(x \notin Q\), is not the pointwise limit of continuous functions, although this is difficult to show. Prove, however, that \(f\) is a pointwise limit of functions that are themselves pointwise limits of continuous functions themselves.

We say that a sequence of functions \(f_{n:} \mathbb{R} \rightarrow \mathbb{R}\) converges uniformly on compact subsets if for every \(k \in \mathbb{N}\), the sequence \(\left\\{f_{n}\right\\}\) comverges uniformly on \(\left.\mid-k, k\right] .\) a) Prove that if \(f_{n}: \mathbb{R} \rightarrow \mathbb{R}\) is a sequence of continuous functions comverging uniformly on compact subsets, then the limit is continuous. b) Prove that if \(f_{n}: \mathbb{R} \rightarrow \mathbb{R}\) is a sequence of functions Riemann integrable on any closed and bounded interval \([a, b],\) and comerging uniformly on compact subsets to an \(f: \mathbb{R} \rightarrow \mathbb{R},\) then for any interval \([a, b],\) we have \(f \in \mathscr{A}[a, b],\) and \(\int_{a}^{b} f=\lim _{n \rightarrow \infty} \int_{a}^{b} f_{n}\)
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