Question

Let \(f(x)=x^{-1 / 2}\) if \(0

Short answer

g is in L¹(m), discontinuous and unbounded everywhere, and g² is finite a.e. but not integrable.

Step-by-step solution

Define the Function f(x)

The function is defined as \( f(x) = x^{-1/2} \) for \( 0 < x < 1 \) and \( f(x) = 0 \) otherwise. This function is only nonzero in the interval \((0, 1)\). This is an important property as it gives us the foundational behavior of \( g(x) \).

Understanding the Sum g(x)

The function \( g(x) \) is constructed by summing over the shifted versions of \( f \) using all rationals, weighted by \( 2^{-n} \). Therefore, \( g(x) = \sum_{n=1}^{\infty} 2^{-n} f(x-r_n) \), where each \( f(x-r_n) \) contributes only when \( x-r_n \) is in \( (0,1) \), i.e., \( r_n < x < r_n+1 \).

Proving g ∈ L¹(m), g

Since the support of each \( f(x - r_n) \) is \((r_n, r_n+1)\) and the sequence of rationals is dense, each point is covered. However, the decay of \( 2^{-n} \) ensures the sum's convergence almost everywhere. Therefore, \( g \in L^{1}(m) \) and \( g<\infty \) a.e., because \( 2^{-n} \) ensures convergence of the series.

Discontinuity and Unboundedness

For every interval \((a, b)\), dense rationals ensure we find rationals arbitrarily close to any point within the interval. Since each term \( f(x-r_n) \) is discontinuous and unbounded locally (as \( f\) is unbounded near zero), \( g(x) \) inherits these properties, i.e., \( g \) is discontinuous and unbounded on every interval, and the dense nature of the rationals prevents a modification that could restrict these properties over a Lebesgue null set.

Squaring g and Integration Issues

The square \( g^2 \) remains finite almost everywhere because the summation method maintains the convergence of squares like the original function. However, \( g^2 \) cannot be integrable over any interval due to elements of \( g(x) \) becoming dominant in any arbitrarily small neighborhood, making their integral infinite.

Key concepts

L¹ Space

The concept of the L¹ space, or Lebesgue integrable functions, is pivotal in functional analysis. A function \( g(x) \in L^1(m) \) means that the integral of the absolute value of \( g \) over a given domain is finite. For our function \( g(x) \), this reinforces that despite its discontinuous nature, it behaves well in terms of integrability.
Locate \( g \) as a sum of functions weighted by \( 2^{-n} \), ensuring that the integral over the absolute values converges.
This convergence is guaranteed by the rapid decay of \( 2^{-n} \), which compensates for possible large values of \( f(x - r_n) \).
In practical terms:

• Check the integral condition: \( \int |g(x)| \, dx < \infty \).
• Use properties like density of rationals to ensure each part of \( g(x) \) is well-covered and contributes appropriately.

Convergence of Series

Convergence is a key mathematical topic where sums of infinite terms behave in a predictable manner. In our function \( g(x) = \sum_{n=1}^{\infty} 2^{-n} f(x-r_n) \), there's particular attention to how the terms converge.
The decay factor \( 2^{-n} \) ensures that each term's contribution diminishes rapidly as \( n \) increases.
This creates a series that converges almost everywhere, given each \( f(x-r_n) \) shifts the non-zero segment of \( f \) across rational-induced intervals.
Key points in ensuring convergence include:

• Continually reducing term size: The exponential decay of \( 2^{-n} \) is critical.
• Dense coverage of the real line: Each \( r_n \) spreads terms across every potential interval.
• Convergence control ensures \( g \) remains finite almost everywhere.

Discontinuity and Unboundedness

Discontinuity and unboundedness are fascinating properties of the function \( g(x) \). These properties indicate that \( g \) has jumps or values that can't be pinned down to finitely behaving intervals.
For \( g \), discontinuity arises because \( f(x-r_n) \), the shifted function, brings points of local discontinuity to all rational intervals.
The dense nature of rationals guarantees that no matter how small an interval is, \( g(x) \) will hit it.
Unbounded behavior roots from \( f \)'s initial definition \( x^{-1/2} \) which spikes near zero.

• This discontinuity makes \( g(x) \) continuously jumpy across its entire domain.
• The unboundedness ensures that \( g(x) \) can have arbitrarily high values within any interval.

This interplay keeps \( g \) from stabilizing under continuous observation.

Rational Numbers and Density

Rational numbers are basic yet incredibly crucial. Their density in the real numbers allows for powerful constructs in functions like \( g(x) \).
Because each rational number \( r_n \) is dense, it serves to shift the function \( f \) continuously across any desired segment of the real line.
This quality makes it possible for \( g(x) \) to remain indefinitely complex and varied across any interval.
Think of these properties:

• Density covers every conceivable point: No real number is without rational neighbors.
• Allows for \( f(x-r_n) \) to enact its changes uniformly across \( x \).
• Provides the fundamental mechanism by which \( g \) inherits its varied behavior.

In simpler terms, rational density lets \( g \) be as unpredictable and vast in behavior throughout its domain.