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Chapter 1: Problem 10

Let \(V\) denote the set of all differentiable real-valued functions defined on the real line. Prove that \(V\) is a vector space with the operations of addition and scalar multiplication defined in Example \(3 .\)

Short Answer

Expert verified
We need to verify the following properties to show that \(V\) is a vector space: 1. Closure under addition: \(f(x) + g(x)\) is a differentiable real-valued function defined on the real line. 2. Closure under scalar multiplication: \(kf(x)\) is a differentiable real-valued function defined on the real line. 3. Commutativity: \(f(x) + g(x) = g(x) + f(x)\). 4. Associativity: \((f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x))\). 5. Identity element for addition: There exists a function \(0(x)\) such that \(f(x) + 0(x) = f(x)\). 6. Inverses for addition: There exists a function \(-f(x)\) such that \(f(x) + (-f(x)) = 0(x)\). 7. Distributive laws for scalars and elements: \(a(f(x) + g(x)) = af(x) + ag(x)\) and \((a + b)f(x) = af(x) + bf(x)\). After verifying all these properties using the properties of differentiable functions, scalar multiplication, and function addition, we can conclude that \(V\) is a vector space with the given operations of addition and scalar multiplication.

Step by step solution

01

Closure under addition

Let \(f(x)\) and \(g(x)\) be two differentiable real-valued functions defined on the real line. Then \(f(x) + g(x)\) is a function defined by the sum of the values of \(f(x)\) and \(g(x)\). We need to show that this function is also a differentiable real-valued function defined on the real line. From the properties of differentiable functions, we know that the derivative of the sum is the sum of the derivatives: \(\frac{d}{dx}[f(x) + g(x)] = \frac{df(x)}{dx} + \frac{dg(x)}{dx}\). Since \(f(x)\) and \(g(x)\) are differentiable functions on the real line, the sum of the derivatives also exists for all real x. Therefore, with the existence of the derivatives, we can conclude that \(f(x) + g(x)\) is also a differentiable real-valued function defined on the real line.
02

Closure under scalar multiplication

Let \(f(x)\) be a differentiable real-valued function defined on the real line and \(k\) be a scalar. Then \(kf(x)\) is a function defined by the scalar multiplication of the values of \(f(x)\). To show that \(kf(x)\) is also a differentiable real-valued function defined on the real line, we must verify that the derivative of this function exists for every real \(x\). We use the properties of differentiable functions and scalar multiplication to do this. We have \(\frac{d}{dx}[kf(x)] = k\frac{df(x)}{dx}\). Since \(f(x)\) is differentiable on the real line, its derivative exists for every real \(x\), then the scalar times its derivative also exists and is a differentiable function for every real value of \(x\). Therefore, \(kf(x)\) is a differentiable real-valued function defined on the real line.
03

Commutativity

For any differentiable functions \(f(x)\) and \(g(x)\), we must show that \(f(x) + g(x) = g(x) + f(x)\). We know that addition of functions is commutative by definition. Thus, for every real \(x\), we have: \(f(x) + g(x) = g(x) + f(x)\).
04

Associativity

For differentiable functions \(f(x)\), \(g(x)\), and \(h(x)\), we must confirm that \((f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x))\). The addition of functions is associative by definition. Thus, for every real \(x\), we have: \((f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x))\).
05

Identity element for addition

An identity element in a vector space is an element that does not change other elements under the operation of addition. We must show that there exists a function \(0(x)\) such that \(f(x) + 0(x) = f(x)\) for every differentiable function \(f(x)\). Let \(0(x)\) be a function defined by \(0(x) = 0\) for every \(x\) in the real line. Since the derivative of a constant function is zero, \(0(x)\) is a differentiable function on the real line. So, for every differentiable function \(f(x)\) and every \(x\), we have: \(f(x) + 0(x) = f(x) + 0 = f(x)\).
06

Inverses for addition

For each differentiable function \(f(x)\), we need to show that there exists a differentiable function \(-f(x)\) such that \(f(x) + (-f(x)) = 0(x)\), where \(0(x)\) is the identity function defined in Step 5. Let \(-f(x)\) be a function defined by \(-f(x) = -1 \cdot f(x)\) for every \(x\) in the real line. We have, for each real \(x\): \(f(x) + (-f(x)) = f(x) - f(x) = 0 = 0(x)\). Since the function \(-f(x)\) is a scalar multiplication of a differentiable function \(f(x)\), it is also a differentiable function on the real line according to Step 2.
07

Distributive laws for scalars and elements

For the distributive laws, we must prove the following for differentiable functions \(f(x), g(x)\) and all scalars \(a, b\): 1. \(a(f(x) + g(x)) = af(x) + ag(x)\). 2. \((a + b)f(x) = af(x) + bf(x)\). 1. For the first law, we have: \(a(f(x) + g(x)) = a[f(x) + g(x)] = af(x) + ag(x)\). 2. For the second law, we have: \((a + b)f(x) = (a + b)f(x) = af(x) + bf(x)\). With each of the properties verified, we can conclude that \(V\), the set of all differentiable real-valued functions defined on the real line, is a vector space using the given operations of addition and scalar multiplication.

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Let \(u, v\), and \(w\) be distinct vectors of a vector space \(\mathrm{V}\). Show that if \(\\{u, v, w\\}\) is a basis for \(\mathrm{V}\), then $\\{u+v+w, v+w, w\\}\( is also a basis for \)\mathrm{V}$.

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