/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 166 Consider the sequence of functio... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 166

Consider the sequence of functions \(\left(f_{n}\right)\) given by $$f_{n}(x)=\left\\{\begin{array}{ll} n & \text { if } x \in\left(0, \frac{1}{n}\right) \\ 0 & \text { otherwise } \end{array}\right.$$ (a) Show that \(f_{n} \stackrel{\text { ptwise }}{\longrightarrow} 0\) on [0,1] , but \(\lim _{n \rightarrow \infty} \int_{x=0}^{1} f_{n}(x) \mathrm{d} x \neq \int_{x=0}^{1} 0 \mathrm{~d} x\). (b) Can the convergence be uniform? Explain.

Short Answer

Expert verified
The sequence of functions \(f_n(x)\) converges pointwise but not uniformly to the zero function on [0,1], and while the integral of the pointwise limit function \(f(x) = 0\) over [0,1] is 0, the limit of integrals of \(f_n(x)\) over [0,1] is 1.

Step by step solution

01

Establish Pointwise Convergence

Pointwise convergence on an interval means that for any given \(x\) in that interval, the sequence of function values \(f_n(x)\) converges as \(n \rightarrow \infty\). Let鈥檚 consider any \(x\) in [0,1]. For \(x = 0\) and \(x > 1/n\), for all \(n > 1/x\) (notice that we can choose such \(n\) because for any \(x > 0 \in [0,1]\), there exists \(N\) such that when \(n > N\), \(1/n < x\)), \(f_n(x) = 0\). As \(n\) becomes larger and larger, \(f_n(x)\) will eventually constantly be 0, even though initially, when \(n \leq 1/x\), \(f_n(x) = n\). Therefore, \(f_n\) converges pointwise to the zero function on [0,1].
02

Show that Limit of Integrals is not Integral of Limit Function

Firstly, let鈥檚 compute the integral of the limit function \(0\) from \(0\) to \(1\), which is \(0\). Now let鈥檚 compute limit of the integrals of \(f_n\) as \(n \rightarrow \infty\). On the interval (0, 1/n), \(f_n(x) = n\), and \(0\) otherwise. So, \(\int_{x=0}^{1} f_{n}(x) dx = \int_{x=0}^{1/n} n dx = n * 1/n = 1\). The limit as \(n \rightarrow \infty\) of this value is \(1\), which is not equal to the integral of the limit function \(0\). Thus, \(\lim _{n \rightarrow \infty} \int_{x=0}^{1} f_{n}(x) dx \neq\int_{x=0}^{1} 0 dx\).
03

Determine if Convergence is Uniform

For uniform convergence of a sequence of functions, given any positive number \(\varepsilon\), there must be \(N\) so that for all \(n \geq N\) and for all \(x\) in the interval, the absolute difference between \(f_n(x)\) and its limit function \(f(x)\) is less than \(\varepsilon\). Here, we know that the limit function \(f(x) = 0\). For each \(n\), there exists \(x\) (specifically when \(x = 1/(2n)\)) such that |\(f_n(x) - 0| = n / 2 > \varepsilon\), no matter how large \(n\) is. Therefore, the convergence cannot be uniform.

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Key Concepts

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

Uniform Convergence
Uniform convergence is a stronger form of convergence for sequences of functions than pointwise convergence. In uniform convergence, a sequence of functions \(f_n(x)\) converges to a function \(f(x)\) in such a way that the speed of convergence is the same across the entire interval. This means that for any chosen small positive number \(\varepsilon\), there exists a number \(N\) such that for all \(n \geq N\) and for all \(x\) in the domain, the absolute difference between the functions, \(|f_n(x) - f(x)|\), is less than \(\varepsilon\).
In the example given in the problem, the sequence of functions \(f_n(x)\) converges to the zero function, but not uniformly. This is because there is always some \(x\) in the interval \((0, \frac{1}{n})\) where \(|f_n(x) - 0| = n > \varepsilon\). Hence, by crafting a carefully chosen \(x\), like \(x = 1/(2n)\), the convergence fails to meet the uniform convergence criteria.
Limit of Integrals
The concept of the "Limit of Integrals" versus "Integral of Limit" is a rich topic in real analysis. The problem highlights an important example where these concepts diverge, indicating a failure of the interchangeability of limit and integration. When dealing with a sequence of functions \(f_n(x)\), an integral of each function across a specified domain might have a limit that differs from the integral of the limit function over the same domain.
In this specific case, for each \(n\), the integral \(\int_{0}^{1} f_{n}(x) \, dx \) evaluates to \(1\). However, the integral of the pointwise limit function, which is \(0\), results in \(0\). As \(n\) approaches infinity, the limit of the integral values is not equal to the integral of the limit function. This discrepancy underscores why pointwise convergence does not guarantee interchangeability between limits and integrals.
Sequence of Functions
A sequence of functions is simply a list of functions ordered in a sequential manner, typically indexed by positive integers \(n\), i.e., \(f_1(x), f_2(x), \ldots, f_n(x)\). These sequences exhibit convergence behaviors, and understanding these behaviors is vital in analyzing their long-term behavior.
For pointwise convergence, as seen in the exercise, we consider the behavior of each function at a fixed point \(x\) as \(n\) becomes large. Specifically, \(f_n(x)\) converges to zero for each individual \(x\), but not uniformly.
  • The sequence converges pointwise to the zero function.
  • It highlights the subtle distinctions between different notions of convergence.
  • It's essential to explore these concepts to grasp fully how sequences of functions behave under various operations, such as integration.
Understanding the nature of a sequence of functions from pointwise to uniform convergence prepares one for deeper mathematical challenges.

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Most popular questions from this chapter

Let \(0

Since the convergence of Cauchy sequences can be taken as the completeness axiom for the real number system, it does not hold for the rational number system. Give an example of a Cauchy sequence of rational numbers which does not converge to a rational number. If we apply the above ideas to series we obtain the following important result, which will provide the basis for our investigation of power series.

Consider the sequence of functions \(\left(f_{n}\right)\) defined on [0,1] by \(f_{n}(x)=x^{n} .\) Show that the sequence converges to the function $$f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x \in[0,1) \\ 1 & \text { if } x=1 \end{array}\right.$$ pointwise on \([0,1],\) but not uniformly on \([0,1] .\) Notice that for the Fourier series at the beginning of this chapter, $$f(x)=\frac{4}{\pi}\left(\cos \left(\frac{\pi}{2} x\right)-\frac{1}{3} \cos \left(\frac{3 \pi}{2} x\right)+\frac{1}{5} \cos \left(\frac{5 \pi}{2} x\right)-\frac{1}{7} \cos \left(\frac{7 \pi}{2} x\right)+\cdots\right)$$ the convergence cannot be uniform on \((-\infty, \infty)\), as the function \(f\) is not continuous. This never happens with a power series, since they converge to continuous functions whenever they converge. We will also see that uniform convergence is what allows us to integrate and differentiate a power series term by term.

Prove Theorem 35. [Hint: \(\left|s_{m}-s_{n}\right|=\left|s_{m}-s+s-s_{n}\right| \leq\) \(\left.\left|s_{m}-s\right|+\left|s-s_{n}\right| \cdot\right]\) So any convergent sequence is automatically Cauchy. For the real number system, the converse is also true and, in fact, is equivalent to any of our completeness axioms: the NIP, the Bolzano-Weierstrass Theorem, or the LUB Property. Thus, this could have been taken as our completeness axiom and we could have used it to prove the others. One of the most convenient ways to prove this converse is to use the Bolzano-Weierstrass Theorem. To do that, we must first show that a Cauchy sequence must be bounded. This result is reminiscent of the fact that a convergent sequence is bounded (Lemma 2 of Chapter 4\()\) and the proof is very similar.

Prove Theorem 34. $$\int_{t=a}^{x} f_{n}^{\prime}(t) \mathrm{d} t=f_{n}(x)-f_{n}(a)$$ Take the limit of both sides and differentiate with respect to \(x .]\) As before, applying this to power series gives the following result.
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