/*! 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 16 Let \(C^{n}(R)\) denote the set ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 16

Let \(C^{n}(R)\) denote the set of all real-valued functions defined on the real line that have a continuous \(n\)th derivative. Prove that \(C^{n}(R)\) is a subspace of \(\mathcal{F}(R, R)\).

Short Answer

Expert verified
To prove that \(C^n(R)\) is a subspace of \(\mathcal{F}(R, R)\), we need to show that \(C^n(R)\) is non-empty and closed under addition and scalar multiplication. We can demonstrate non-emptiness using the constant function \(f(x)=0\). We show closure under addition by taking \(f(x), g(x) \in C^n(R)\) and proving that their sum, \(h(x)=f(x)+g(x)\), belongs to \(C^n(R)\) as \((h(x))^{(n)} = f^{(n)}(x) + g^{(n)}(x)\) is continuous. Similarly, for closure under scalar multiplication, we take \(f(x) \in C^n(R)\) and a scalar \(k\), then the function \(l(x)=kf(x)\) is also in \(C^n(R)\) since \((l(x))^{(n)} = k(f^{(n)}(x))\) is continuous. Thus, \(C^n(R)\) is a subspace of \(\mathcal{F}(R, R)\).

Step by step solution

01

1. Non-emptiness

To show non-emptiness, we must show that there exists at least one function in \(C^n(R)\). A simple example would be the constant function \(f(x) = 0\), as all of its derivatives will also be zero and hence continuous. Therefore, \(C^n(R)\) is not empty.
02

2. Closure under addition

Next, we need to show that if \(f(x)\) and \(g(x)\) belong to \(C^n(R)\), then \(h(x) = f(x) + g(x)\) should also belong to \(C^n(R)\). Since both \(f(x)\) and \(g(x)\) have continuous nth derivatives, we will compute the nth derivative of \(h(x)\) and show that it is continuous: \((h(x))^{(n)} = (f(x) + g(x))^{(n)} = f^{(n)}(x) + g^{(n)}(x)\) The sum of two continuous functions is also continuous. Therefore, \(h^{(n)}(x)\) is continuous, and \(h(x)\) belongs to \(C^n(R)\). This proves closure under addition.
03

3. Closure under scalar multiplication

Lastly, we need to show that if \(f(x)\) belongs to \(C^n(R)\) and \(k\) is a scalar, then \(l(x) = kf(x)\) should also belong to \(C^n(R)\). We will compute the nth derivative of \(l(x)\) and show that it is continuous: \((l(x))^{(n)} = (kf(x))^{(n)} = k(f^{(n)}(x))\) The product of a scalar and a continuous function is also continuous. Therefore, \(l^{(n)}(x)\) is continuous, and \(l(x)\) belongs to \(C^n(R)\). This proves closure under scalar multiplication. Since \(C^n(R)\) satisfies the non-emptiness property and is closed under addition and scalar multiplication, we can conclude that \(C^n(R)\) is a subspace of \(\mathcal{F}(R, R)\).

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