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Prove that if \(f: S \rightarrow T\) is uniformly continuous on \(B \subseteq S,\) and \(g: T \rightarrow U\) is uniformly continuous on \(f[B]\), then the composite function \(g \circ f\) is uniformly continuous on \(B\).

Short Answer

Expert verified
The composite \(g \circ f\) is uniformly continuous on \(B\).

Step by step solution

01

Define Uniform Continuity

A function \(h: A \rightarrow B\) is uniformly continuous if, for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y \in A\), if \(||x - y|| < \delta\), then \(||h(x) - h(y)|| < \epsilon\).
02

Apply Uniform Continuity to f

Since \(f: S \rightarrow T\) is uniformly continuous on \(B \subseteq S\), given \(\epsilon > 0\), there exists a \(\delta_1 > 0\) so that for all \(x, y \in B\), if \(||x - y|| < \delta_1\) then \(||f(x) - f(y)|| < \frac{\epsilon}{2}\).
03

Apply Uniform Continuity to g

Since \(g: T \rightarrow U\) is uniformly continuous on \(f[B]\), given \(\epsilon > 0\), there exists a \(\delta_2 > 0\) so that for all \(u, v \in f[B]\), if \(||u - v|| < \delta_2\) then \(||g(u) - g(v)|| < \epsilon\).
04

Combine the Conditions for g and f

Utilize the \(\delta_1\) from Step 2 and the \(\delta_2\) from Step 3. If for \(x, y \in B\), \(||x - y|| < \delta_3\) where \(\delta_3 = \min(\delta_1, \delta_2)\), then \(||f(x) - f(y)|| < \frac{\epsilon}{2}\), and since \(f(x), f(y) \in f[B]\) as \(||f(x) - f(y)|| < \delta_2\), we have \(||g(f(x)) - g(f(y))|| < \epsilon\).
05

Conclude Uniform Continuity of Composite

Therefore, for \(g \circ f\), given \(\epsilon > 0\), there exists \(\delta_3 > 0\) such that \(||x - y|| < \delta_3\) leads to \(||g(f(x)) - g(f(y))|| < \epsilon\), thus proving \(g \circ f\) is uniformly continuous on \(B\).

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Key Concepts

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

Composite Functions
When dealing with functions, especially in advanced mathematics, the idea of composite functions is crucial. A composite function is formed when one function is applied to the results of another. In mathematical terms, if we have two functions, say \(f\) and \(g\), the composite function \(g \circ f\) applies function \(f\) first and then \(g\) on the resulting values.
This can be symbolized as:
  • \(f: S \rightarrow T\)
  • \(g: T \rightarrow U\)
  • \(g \circ f: S \rightarrow U\)
This means that the output of \(f\) becomes the input for \(g\). Understanding composite functions is vital when analyzing continuity and other properties through consecutive transformations, as the behavior of the output depends on both functions acting in tandem.
In the exercise at hand, we are looking at how the uniform continuity of individual functions \(f\) and \(g\) ensures the uniform continuity of the composite \(g \circ f\). This is a foundational concept in mathematical analysis, often serving as a stepping stone to more intricate properties like differentiability and integrals.
Epsilon-Delta Definition
The epsilon-delta definition is a fundamental part of real analysis and calculus, used to rigorously define the notion of continuity. This concept ties closely with both pointwise and uniform continuity. To say that a function \(h: A \rightarrow B\) is uniformly continuous means that the distance between outputs \(h(x)\) and \(h(y)\) can be made arbitrarily small whenever the distance between \(x\) and \(y\) is sufficiently small.
The definition goes like this:
  • For every \(\epsilon > 0\) given (which represents how close the function outputs need to be),
  • there exists a \(\delta > 0\) (representing how close the inputs need to be) such that for all points \(x, y\) in the domain of \(h\),
  • if \(||x - y|| < \delta\), then \(||h(x) - h(y)|| < \epsilon\).
This uniform condition must hold for the entire domain of \(h\), making it stronger than regular continuity where \(\delta\) may depend on the point \(x\) in question. The exercise illustrates applying this principle to prove that a composite function \(g \circ f\) retains uniform continuity when both \(f\) and \(g\) possess this trait.
Uniform Convergence
Uniform convergence is another important concept in analysis, often considered alongside uniform continuity. It's a type of convergence related to sequences or series of functions. The importance of uniform convergence comes into play during the approximation of functions and ensuring properties like integration and differentiation are preserved through the limit process.
For example, if a sequence of functions \(\{f_n(x)\}\) converges to a function \(f(x)\) uniformly on some set, it implies:
  • For any \(\epsilon > 0\), there exists an integer \(N\) such that for all \(n \geq N\),
  • and for all \(x\) in the set, \(||f_n(x) - f(x)|| < \epsilon\).
The distinguishing factor is the independence of \(x\) in choosing \(N\), maintaining the closeness of \(f_n(x)\) to \(f(x)\) uniformly over the entire set. Although uniform convergence is more about sequences rather than single functions being continuous, understanding it provides deeper insight into how the behavior of functions can be controlled and predicted over entire intervals or domains.
In the context of our exercise, uniform continuity and uniform convergence both share the idea of control over entire domains, helping ensure smooth and consistent transformations in functions, especially when analyzing compositional behavior.

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Most popular questions from this chapter

Prove, independently, the principle of nested intervals in \(E^{n},\) i.e., Theorem 5 with $$ F_{m}=\left[\bar{a}_{m}, \bar{b}_{m}\right] \subseteq E^{n}; $$ where $$ \bar{a}_{m}=\left(a_{m 1}, \ldots, a_{m n}\right) \text { and } \bar{b}_{m}=\left(b_{m 1}, \ldots, b_{m n}\right) $$ [Hint: As \(F_{m+1} \subseteq F_{m}, \bar{a}_{m+1}\) and \(\bar{b}_{m+1}\) are in \(F_{m} ;\) hence by properties of closed intervals, \(a_{m k} \leq a_{m+1, k} \leq b_{m+1, k} \leq b_{m k}, \quad k=1,2, \ldots, n\) Fixing \(k\), let \(A_{k}\) be the set of all \(a_{m k}, m=1,2, \ldots\) Show that \(A_{k}\) is bounded above by each \(b_{m k},\) so let \(p_{k}=\sup A_{k}\) in \(E^{1}\). Then $$ (\forall m) \quad a_{m k} \leq p_{k} \leq b_{m k} . \text { (Why?) } $$ Unfixing \(k\), obtain such inequalities for \(k=1,2, \ldots, n\). Let \(\bar{p}=\left(p_{1}, \ldots, p_{k}\right)\). Then $$ (\forall m) \quad \bar{p} \in\left[\bar{a}_{m}, \bar{b}_{m}\right], \text { i.e., } \bar{p} \in \bigcap F_{m}, \text { as required. } $$ Note that the theorem fails for nonclosed intervals, even in \(E^{1} ;\) e.g., take \(F_{m}=\) \((0,1 / m]\) and show that \(\bigcap_{m} F_{m}=\emptyset\).]

Let \(\bar{p}_{0}, \bar{p}_{1}, \ldots, \bar{p}_{m}\) be fixed points in \(E^{n}\left({ }^{*}\right.\) or in another normed space \()\). Let $$f(t)=\bar{p}_{k}+(t-k)\left(\bar{p}_{k+1}-\bar{p}_{k}\right)$$ whenever \(k \leq t \leq k+1, t \in E^{1}, k=0,1, \ldots, m-1\) Show that this defines a uniformly continuous mapping \(f\) of the interval \([0, m] \subseteq E^{1}\) onto the "polygon" $$\bigcup_{k=0}^{m-1} L\left[p_{k}, p_{k+1}\right]$$ In what case is \(f\) one to one? Is \(f^{-1}\) uniformly continuous on each \(L\left[p_{k}, p_{k+1}\right] ?\) On the entire polygon?

A set is said to be totally disconnected iff its only connected subsets are one-point sets and \(\emptyset\). Show that \(R\) (the rationals) has this property in \(E^{1}\).

Prove that if \(X\) and \(Y\) are connected, so is \(X \times Y\) under the product metric.

Prove that \(A\) is connected iff there is no continuous map $$ f: A \underset{\text { onto }}{\longrightarrow}\\{0,1\\} .^{5} $$ [Hint: If there is such a map, Theorem 1 shows that \(A\) is disconnected. (Why?) Conversely, if \(A=P \cup Q(P, Q\) as in Definition 3), put \(f=0\) on \(P\) and \(f=1\) on \(Q\). Use again Theorem 1 to show that \(f\) so defined is continuous on \(A\).]

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