Suppose 𝑏 ∈ 𝐑. Show that the set of continuous real-valued functions 𝑓 on
the interval [0, 1] such that ∫
1
0
𝑓 = 𝑏 is a subspace of 𝐑^[0,1] if and only if
𝑏 = 0.

To prove the statement, we need to show that the set of continuous real-valued functions \( f \) on the interval \([0, 1]\) such that 

\[
\int_0^1 f(x) \, dx = b
\]

is a subspace of the vector space \( \mathbb{R}^{[0,1]} \) (the space of all continuous real-valued functions on \([0, 1]\)) **if and only if** \( b = 0 \).

### Definitions and Conditions for a Subspace:
A subset \( V \) of a vector space \( W \) is a subspace if it satisfies the following conditions:
1. **Zero Vector:** \( V \) contains the zero vector of \( W \).
2. **Closed under Addition:** If \( f, g \in V \), then \( f + g \in V \).
3. **Closed under Scalar Multiplication:** If \( f \in V \) and \( c \in \mathbb{R} \), then \( c f \in V \).

Let \( V = \{ f \in \mathbb{R}^{[0,1]} \mid \int_0^1 f(x) \, dx = b \} \).

#### (⇐) **Suppose \( b = 0 \).**
If \( b = 0 \), then \( V = \{ f \in \mathbb{R}^{[0,1]} \mid \int_0^1 f(x) \, dx = 0 \} \).

1. **Zero Vector:** 
   The zero function \( f(x) = 0 \) for all \( x \in [0, 1] \) is in \( V \) because
   \[
   \int_0^1 0 \, dx = 0.
   \]
   Hence, the zero vector is in \( V \).

2. **Closed under Addition:**
   Let \( f, g \in V \). Then,
   \[
   \int_0^1 f(x) \, dx = 0 \quad \text{and} \quad \int_0^1 g(x) \, dx = 0.
   \]
   Therefore,
   \[
   \int_0^1 (f(x) + g(x)) \, dx = \int_0^1 f(x) \, dx + \int_0^1 g(x) \, dx = 0 + 0 = 0.
   \]
   Thus, \( f + g \in V \), so \( V \) is closed under addition.

3. **Closed under Scalar Multiplication:**
   Let \( f \in V \) and \( c \in \mathbb{R} \). Then,
   \[
   \int_0^1 f(x) \, dx = 0.
   \]
   Therefore,
   \[
   \int_0^1 (c f(x)) \, dx = c \int_0^1 f(x) \, dx = c \cdot 0 = 0.
   \]
   Thus, \( c f \in V \), so \( V \) is closed under scalar multiplication.

Since all three conditions for a subspace are satisfied, \( V \) is a subspace when \( b = 0 \).

#### (⇒) **Suppose \( V \) is a subspace. We must show that \( b = 0 \).**

Assume for contradiction that \( b \neq 0 \).

1. **Zero Vector Condition:**
   The zero vector (the zero function) must be in \( V \). So, 
   \[
   \int_0^1 0 \, dx = 0 \quad \text{must hold, implying that \( b = 0 \), contradicting \( b \neq 0 \)}.
   \]

2. If \( b \neq 0 \), consider \( f(x) \) such that \( \int_0^1 f(x) \, dx = b \). Since \( V \) is a subspace, \( 0 \cdot f(x) = 0 \) (the zero function) must also be in \( V \). This would imply \( \int_0^1 0 \, dx = b \), which again forces \( b = 0 \), a contradiction.

Thus, \( V \) cannot be a subspace unless \( b = 0 \).

### Conclusion:
The set \( \{ f \in \mathbb{R}^{[0,1]} \mid \int_0^1 f(x) \, dx = b \} \) is a subspace of \( \mathbb{R}^{[0,1]} \) **if and only if** \( b = 0 \).

---

Do you have any questions or want further details? Here are some related questions:

1. What is the definition of a vector space?
2. How do we prove that a set is not a subspace?
3. Can you provide an example of a non-zero \( b \) where the set is not a subspace?
4. How does scalar multiplication affect the integral condition?
5. What are the implications of this result in functional analysis?

**Tip:** When dealing with subspaces, always check the three fundamental conditions: zero vector, closure under addition, and closure under scalar multiplication.

Explore the proof that the set of continuous real-valued functions on [0, 1] with the integral equal to b forms a subspace of the vector space of all continuous functions if and only if b equals zero. Includes detailed mathematical reasoning and examples.

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