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Chapter 10: Problem 10

If \(f, g \in \mathcal{L}[a, b]\) and if \(g\) is bounded and monotone, show that \(f g \in \mathcal{L}[a, b]\). More exactly, if \(|g(x)|\) \(\leq B\), show that \(\|f g\| \leq B\|f\|\).

Short Answer

Expert verified
Product of an integrable function and a bounded and monotone function is integrable and satisfies the property \( \|f g\| \leq B \|f\| \).

Step by step solution

01

Understand bounded and monotone functions

A function g is considered bounded if there exists a number B such that \( |g(x)| <= B \) for all x. This essentially means that the function never exceeds B in absolute value. A function g is considered monotone if it is either entirely non-increasing, or entirely non-decreasing.
02

Show that \( f \cdot g \) is integrable

To handle the multiplication of the two functions \( f \cdot g \), we can break it down as follows: Let the function \( f \) be integrable and the function \( g \) be a bounded function. It is given that \( |g(x)| \leq B \), where B is a constant. So, the absolute value of the product \( f(x) \cdot g(x) \) is \( |f(x) \cdot g(x)| = |f(x)| \cdot |g(x)| \leq |f(x)| \cdot B = B|f(x)| \). Since \( f \) is integrable in the specific interval [a, b], \( f \cdot g \) must also be integrable in the interval [a, b].
03

Proving the Bounded Property

The norm \( \|f g\| \) of the function \( f \cdot g \) is defined as the absolute integral \( \int |f g| dx \) in the interval [a, b]. Now, from step 2, we know that \( |f g| \leq B |f| \). Hence, \( \int |f g| dx \leq \int B |f| dx = B \int |f| dx = B \|f\| \). So, it follows that \( \|f g\| \leq B \|f\| \)

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Key Concepts

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

Bounded Functions
Imagine you're walking your dog with a leash that's only so long. No matter how excited your furry friend gets, they can't sprint beyond the length of the leash. Now, let's think of a mathematical function in a similar way. A function is called bounded if it's tied to a leash of a specific length, figuratively speaking. Formally, a function \( g(x) \) is bounded on an interval if there's a number \( B \) such that the absolute value of \( g(x) \) never exceeds \( B \) within that interval. That's why we say \( |g(x)| \leq B \).
  • A bounded function never 'runs off' to infinity.
  • The number \( B \) acts as the limit for the function's values.
  • In practical terms, bounded functions are predictable and confined within a range.
Understanding this concept is crucial as it indicates that bounded functions are 'tame' in a sense. They have a maximum and a minimum value within an interval. When you multiply a bounded function by an integrable function, the result is also 'tame,' which is an essential step in solving our exercise.
Monotone Functions
Much like a one-way road where traffic flows in only one direction, monotone functions have a similar behavior. Monotone functions are functions that are either always increasing or always decreasing. These functions don鈥檛 have to increase or decrease rapidly; they just need to stick to one direction throughout their domain. Monotone functions are predictable because they don't suddenly change course.

Properties of Monotone Functions:

  • If a function is increasing, it means for any two points, if \( x_1 < x_2 \), then \( g(x_1) \leq g(x_2) \).
  • If a function is decreasing, it鈥檚 just the opposite: \( g(x_1) \geq g(x_2) \).
  • These functions provide a level of control when combined with other functions.
In the context of our exercise, a bounded and monotone function such as \( g \) ensures that when multiplied with \( f \)鈥攑rovided \( f \) is integrable鈥攖he resulting function maintains integrability. This is partly because the monotonicity of \( g \) prevents any erratic behavior in the product, making the analysis and conclusions about integrability straightforward.
Absolute Integral
Think of the absolute integral as the total measure of a function's landscape over an interval, ignoring the dips below the ground level. When we calculate the absolute integral of a function, we're considering the area under the curve of the function's absolute value, effectively treating all negative parts as if they were positive.

Implications of the Absolute Integral:

  • It quantifies the overall 'magnitude' of a function on an interval, disregarding the sign.
  • The notation for the absolute integral of \( f \) on \( [a, b] \) is \( \|f\| \) and it equals \( \int_a^b |f(x)| dx \).
  • Assessing the absolute integral helps in determining the bounded nature of a function's product with another function.
In our exercise, by showing that the absolute integral of \( f \) multiplied by a bounded function \( g \) 鈥� \( \int |f g| dx \) 鈥� is less than or equal to \( B \int |f| dx \), we effectively demonstrate that the product \( f \) and \( g \) is not only bounded but has its boundaries firmly established by \( B \) times the absolute integral of \( f \) itself. This is an insightful step in the mathematical journey of understanding the interplay between functions, their limits, and integrals.

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Most popular questions from this chapter

Show that the function \(g_{1}(x):=x^{-1 / 2} \sin (1 / x)\) for \(x \in(0,1]\) and \(g_{1}(0):=0\) belongs to \(\mathcal{R}^{*}[0,1] .\left[\right.\) Hint \(:\) Differentiate \(C_{1}(x):=x^{3 / 2} \cos (1 / x)\) for \(x \in(0,1]\) and \(\left.C_{1}(0):=0 .\right]\)

If \(\delta\) is defined on \([0,2]\) by \(\delta(t):=\frac{1}{2}|t-1|\) for \(x \neq 1\) and \(\delta(1):=0.01\), show that every \(\delta\) -fine partition \(\dot{\mathcal{P}}\) of \([0,2]\) has \(t=1\) as a tag for at least one subinterval, and that the total length of the subintervals in \(\dot{\mathcal{P}}\) having 1 as a tag is \(\leq 0.02\).

Show that the following functions belong to \(\mathcal{R}^{*}[0,1]\) by finding a function \(F_{k}\) that is continuous on \([0,1]\) and such that \(F_{k}^{\prime}(x)=f_{k}(x)\) for \(x \in[0,1] \backslash E_{k}\), for some finite set \(E_{k}\). (a) \(f_{1}(x):=(x+1) / \sqrt{x}\) for \(x \in(0,1]\) and \(f_{1}(0):=0\). (b) \(f_{2}(x):=x / \sqrt{1-x}\) for \(x \in[0,1)\) and \(f_{2}(1):=0\). (c) \(f_{3}(x):=\sqrt{x} \ln x\) for \(x \in(0,1]\) and \(f_{3}(0):=0\). (d) \(f_{4}(x):=(\ln x) / \sqrt{x}\) for \(x \in(0,1]\) and \(f_{4}(0):=0\). (e) \(f_{5}(x):=\sqrt{(1+x) /(1-x)}\) for \(x \in[0,1)\) and \(f_{5}(1):=0\). (f) \(f_{6}(x):=1 /(\sqrt{x} \sqrt{2-x})\) for \(x \in(0,1]\) and \(f_{6}(0):=0\).

(a) If \(\left(f_{k}\right)\) is a bounded sequence in \(\mathcal{M}[a, b]\) and \(f_{k} \rightarrow f\) a.e., show that \(f \in \mathcal{M}[a, b]\). [Hint: Use the Dominated Convergence Theorem.] (b) If \(\left(g_{k}\right)\) is any sequence in \(\mathcal{M}[a, b]\) and if \(f_{k}:=\) Arctan \(\circ g_{k}\), show that \(\left(f_{k}\right)\) is a bounded sequence in \(\mathcal{M}[a, b]\). (c) If \(\left(g_{k}\right)\) is a sequence in \(\mathcal{M}[a, b]\) and if \(g_{k} \rightarrow g\) a.e., show that \(g \in \mathcal{M}[a, b]\).

Let \(f\) and \(|f|\) belong to \(\mathcal{R}^{*}[a, \gamma]\) for all \(\gamma \geq a\). Show that \(f \in \mathcal{L}[a, \infty)\) if and only if for every \(\varepsilon>0\) there exists \(K(\varepsilon) \geq a\) such that if \(q>p>K(\varepsilon)\) then \(\int_{p}^{q}|f|<\varepsilon\)
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