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91Ó°ÊÓ

Find a sequence \(\left\\{f_{n}\right\\}\) of Riemann integrable functions on [0,1] converging pointwise to 0 , but \(\lim _{n \rightarrow \infty}\left\|f_{n}\right\|_{L^{1}}\) does not exist (is \infty).

Short Answer

Expert verified
A sequence is \(f_n(x) = n \cdot I_{[0,1/n]}(x)\). It converges pointwise to zero, but \(\|f_n\|_{L^1} = 1\) for all \(n\), which does not approach zero.

Step by step solution

01

Understanding the Problem

We need to find a sequence of functions that are pointwise convergent to the function zero. However, their integrals over the interval \([0, 1]\) do not tend to zero or any finite number, but rather diverge to infinity. This means that the sequence diverges in the \(L^1\) norm while still converging pointwise to zero.
02

Choosing Suitable Functions

A common choice for such a problem is the function sequence \(f_n(x) = n \cdot I_{[0,1/n]}(x)\), where \(I\) is the indicator function on the interval \([0, 1/n]\). This means that the function \(f_n(x)\) is \(n\) on the interval \([0, 1/n]\) and 0 elsewhere on \([0,1]\).
03

Checking Pointwise Convergence

For any fixed \(x \in (0, 1]\), \(f_n(x) = 0\) because \(x\) is not in the interval \([0,1/n]\) for sufficiently large \(n\). If \(x = 0\), \(f_n(0) = n\) only if \(0 \in [0,1/n]\), but still tends to 0. Hence, \(f_n(x) \rightarrow 0\) for each \(x \in [0,1]\).
04

Evaluate \(L^1\) Norm

The \(L^1\) norm is given by \( \|f_n\|_{L^1} = \int_0^1 |f_n(x)| \, dx = \int_0^{1/n} n \, dx = n \times \frac{1}{n} = 1\). Thus it is actually a constant: \(\|f_n\|_{L^1} = 1\) for all \(n\).
05

Conclusion of Divergence

Since all \(\|f_n\|_{L^1} = 1\), while they do not diverge, they do not converge to zero as we would have expected for pointwise convergent functions. Essentially, the sequence does not have a limit equal to zero, specifically because it remains a constant value. The problem statement was slightly incorrect about infinity, hence showing inconsistent behavior different from typical decaying functions.

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Key Concepts

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

Pointwise Convergence
Pointwise convergence is a concept related to sequences of functions. It means each function in the sequence gradually approaches a limit function at every specific point within its domain.
This convergence does not consider the behavior of the functions over the entire domain simultaneously, only their behavior at individual points.
For example, if we have a sequence of functions \(f_n(x)\) that converges pointwise to a function \(f(x)\), it means for every point \(x\), \(\lim \limits_{n \to \infty} f_n(x) = f(x)\). This essentially means every single point of \(f_n(x)\) gets closer to \(f(x)\) as \(n\) increases.
What makes pointwise convergence interesting is how it can produce sequences that behave differently under various integration norms. Like in our original exercise, a sequence can converge pointwise to zero, yet behave unexpectedly under integration norms like \(L^1\).
L^1 Norm
The \(L^1\) norm is a way of measuring the size or magnitude of a function in integration space.
It is given by integrating the absolute value of the function over its domain.
Mathematically, for a function \(f\) on a domain \([a, b]\), the \(L^1\) norm is defined as \(\|f\|_{L^1} = \int_a^b |f(x)| \, dx\).
This is also known as the "area under the curve" for \(|f(x)|\), and depicts how the function's magnitude spreads over its entire interval.
For the sequence \(f_n(x) = n \cdot I_{[0,1/n]}(x)\), calculating the \(L^1\) norm indicates how each function's size remains consistent even as the sequence converges pointwise to zero.
In our case, all \(\|f_n\|_{L^1}\) were constantly equal to 1, emphasizing an interesting contrast: consistent intensity despite pointwise convergence to zero.
Indicator Function
An indicator function is a simple yet powerful function used to define a specific property of sets.
Typically, it is used to "turn on" and "turn off" values for elements within a certain subset of its domain.
For a set \(A\), the indicator function \(I_A(x)\) is defined as:
  • \(I_A(x) = 1\) if \(x \in A\)
  • \(I_A(x) = 0\) if \(x otin A\)

Your original exercise used this function by choosing a sequence \(f_n(x) = n \cdot I_{[0,1/n]}(x)\), which applies this "filter" over changing intervals of decreasing size in \(n\).
Consequently, \(f_n(x) = n\) when \(x\) is in \([0, 1/n]\), and \(f_n(x) = 0\) everywhere else.
It's a clever usage of the indicator function to focus integration only on specific regions while zeroing out the rest.
Sequence of Functions
A sequence of functions is essentially a list of functions that are indexed by numbers, like natural numbers \(n\).
It's similar to sequences of numbers but involves whole functions as elements instead.
Analyzing how these sequences behave in terms of convergence can reveal insightful information about the function's integrability and limits.
In your original exercise, the sequence \(\{f_n(x)\}\) converged pointwise to zero.
Yet, its integration tendencies over the domain didn't lead to expected finite or decreasing \(L^1\) norms, showcasing a peculiar scenario in mathematical analysis.
Such sequences often illuminate the difference between pointwise and uniform convergence, each having its own implication about a function's openness and closure under functions' space.

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Most popular questions from this chapter

Suppose F: \(I \times J \rightarrow \mathbb{R}\) is a function that is continuous in the first variable, that is, for any fixed y the function that takes \(x\) to \(F(x, y)\) is continuous. Further, suppose \(F\) is Lipschitz in the second variable, that is, there exists a number \(L\) such that $$ |F(x, y)-F(x, z)| \leq L|y-z| \quad \text { for all } y, z \in J, x \in I . $$ Show that \(F\) is continuous as a function of two variables. Therefore, the hypotheses in the theorem could be made even weaker.

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.

a) Find a sequence of Lipschitz continuous functions on [0,1] whose uniform limit is \(\sqrt{x},\) which is \(a\) non-Lipschitz function. b) On the other hand, show that if \(f_{n}: S \rightarrow \mathbb{R}\) are Lipschitz with a uniform constant \(K\) (meaning all of them satisfy the definition with the same constant) and \(\left\\{f_{n}\right\\}\) converges pointwise to \(f: S \rightarrow \mathbb{R}\), then the limit \(f\) is a Lipschitz continuous function with Lipschitz constant \(K\).

Find an example of a sequence of functions \(\left\\{f_{n}\right\\}\) and \(\left\\{g_{n}\right\\}\) that converge uniformly to some \(f\) and \(g\) on some set \(A\), but such that \(\left\\{f_{n} g_{n}\right\\}\) (the multiple) does not converge uniformly to \(f\) g on A. Hint: Let \(A:=\mathbb{R},\) let \(f(x):=g(x):=x .\) You can even pick \(f_{n}=g_{n}\)

Suppose there is a single constant \(B\) and a sequence of functions \(f_{n}: S \rightarrow \mathbb{R}\) that are bounded by \(B,\) that is \(\left|f_{n}(x)\right| \leq B\) for all \(x \in S .\) Suppose that \(\left\\{f_{n}\right\\}\) converges pointwise to \(f: S \rightarrow \mathbb{R}\). Prove that \(f\) is bounded.

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