Question 1166343: 6 Let V = C[a, b] be the vector space of continuous functions on interval
[a, b]. Let W be the subset of functions in V such that:
(integralsign) from(a,b) f(x) dx = 0.
Is W a subspace of V ? You must justify your answer.
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! Yes, **W** is a **subspace** of $\mathbf{V}$, the vector space of continuous functions on the interval $[a, b]$.
The vector space $V$ is $C[a, b]$, the set of all continuous real-valued functions on the interval $[a, b]$. The subset $W$ is defined as:
$$W = \left\{ f(x) \in V \ \bigg| \ \int_a^b f(x) \ dx = 0 \right\}$$
To prove that $W$ is a subspace of $V$, we must verify the three conditions:
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## 1. The Zero Vector Condition (Contains the Zero Function)
The zero vector in $V$ is the **zero function**, $\mathbf{z}(x) = 0$, for all $x \in [a, b]$.
We must check if the zero function satisfies the condition for $W$:
$$\int_a^b \mathbf{z}(x) \ dx = \int_a^b 0 \ dx = 0$$
Since the integral of the zero function is $0$, the zero function is in $W$.
Thus, $W$ is non-empty.
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## 2. Closure under Vector Addition ➕
Let $f(x)$ and $g(x)$ be two arbitrary functions in $W$. This means:
* $\int_a^b f(x) \ dx = 0$
* $\int_a^b g(x) \ dx = 0$
We need to check if their sum, $(f+g)(x) = f(x) + g(x)$, is also in $W$. We check the integral of the sum:
$$\int_a^b (f(x) + g(x)) \ dx = \int_a^b f(x) \ dx + \int_a^b g(x) \ dx \quad \text{(Property of Integrals)}$$
Substitute the conditions for $f$ and $g$:
$$\int_a^b (f(x) + g(x)) \ dx = 0 + 0 = 0$$
Since the integral of the sum is $0$, the function $(f+g)(x)$ is in $W$.
Thus, $W$ is closed under function addition.
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## 3. Closure under Scalar Multiplication ✖️
Let $f(x)$ be an arbitrary function in $W$ (so $\int_a^b f(x) \ dx = 0$), and let $c$ be an arbitrary scalar (a real number).
We need to check if the scalar multiple, $(cf)(x) = c \cdot f(x)$, is also in $W$. We check the integral of the scalar multiple:
$$\int_a^b c \cdot f(x) \ dx = c \cdot \int_a^b f(x) \ dx \quad \text{(Property of Integrals)}$$
Substitute the condition for $f$:
$$\int_a^b c \cdot f(x) \ dx = c \cdot (0) = 0$$
Since the integral of the scalar multiple is $0$, the function $(cf)(x)$ is in $W$.
Thus, $W$ is closed under scalar multiplication.
***
Since $W$ satisfies all three conditions, it is a **subspace** of the vector space $V$.
Would you like to analyze a different set of continuous functions to see if it forms a subspace?
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