Show that if K > 0 and T ∈ B(C([0, 1])) with T(1 ) = 1 , then
T(LipK[0, 1]) is equicontinuous. Here, 1 denotes the constant function with value 1, and LipK[0, 1] denotes
class of K-Lipschitz functions.

To show that if \( K > 0 \) and \( T \in B(C([0, 1])) \) with \( T(1) = 1 \), then \( T(\text{Lip}_K[0, 1]) \) is equicontinuous, where \( 1 \) denotes the constant function with value 1 and \( \text{Lip}_K[0, 1] \) denotes the class of \( K \)-Lipschitz functions, we proceed as follows:

### Definitions and Setup:
1. **Lipschitz functions**: A function \( f: [0,1] \rightarrow \mathbb{R} \) is \( K \)-Lipschitz if there exists a constant \( K > 0 \) such that for all \( x, y \in [0, 1] \),
   \[
   |f(x) - f(y)| \leq K|x - y|.
   \]
   The set of all \( K \)-Lipschitz functions on \([0, 1]\) is denoted by \( \text{Lip}_K[0, 1] \).

2. **Equicontinuity**: A family of functions \( \mathcal{F} \subseteq C([0, 1]) \) is equicontinuous if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( f \in \mathcal{F} \) and for all \( x, y \in [0, 1] \),
   \[
   |x - y| < \delta \implies |f(x) - f(y)| < \epsilon.
   \]

3. **Operators in \( B(C([0, 1])) \)**: The space \( B(C([0, 1])) \) consists of bounded linear operators on the Banach space \( C([0, 1]) \) of continuous functions on \([0, 1]\).

### Step-by-Step Proof:

1. **Condition \( T(1) = 1 \)**: The fact that \( T(1) = 1 \) tells us that \( T \) maps the constant function \( 1 \) (which takes the value 1 everywhere on \([0, 1]\)) to itself.

2. **Action of \( T \) on \( \text{Lip}_K[0, 1] \)**: Consider any function \( f \in \text{Lip}_K[0, 1] \). Since \( T \) is a bounded linear operator on \( C([0, 1]) \), the image \( T(f) \) is also a continuous function on \([0, 1]\).

3. **Lipschitz Condition**: For \( f \in \text{Lip}_K[0, 1] \), we have:
   \[
   |f(x) - f(y)| \leq K |x - y| \quad \text{for all } x, y \in [0, 1].
   \]
   Applying \( T \) to this inequality, and using the linearity of \( T \),
   \[
   |T(f)(x) - T(f)(y)| = |T(f(x) - f(y))| \leq \|T\| \cdot |f(x) - f(y)| \leq \|T\| \cdot K \cdot |x - y|.
   \]
   Thus, \( T(f) \) is \( \|T\|K \)-Lipschitz.

4. **Equicontinuity**: The family \( T(\text{Lip}_K[0, 1]) \) consists of functions \( T(f) \) where \( f \in \text{Lip}_K[0, 1] \). Since each \( T(f) \) is \( \|T\|K \)-Lipschitz, we have for any \( f \in \text{Lip}_K[0, 1] \),
   \[
   |T(f)(x) - T(f)(y)| \leq \|T\|K |x - y|.
   \]
   This implies that the family \( T(\text{Lip}_K[0, 1]) \) is uniformly Lipschitz with Lipschitz constant \( \|T\|K \), which immediately implies equicontinuity.

   Specifically, for any \( \epsilon > 0 \), if we choose \( \delta = \frac{\epsilon}{\|T\|K} \), then for all \( f \in \text{Lip}_K[0, 1] \) and for all \( x, y \in [0, 1] \) with \( |x - y| < \delta \),
   \[
   |T(f)(x) - T(f)(y)| \leq \|T\|K |x - y| < \epsilon.
   \]
   This shows that \( T(\text{Lip}_K[0, 1]) \) is equicontinuous.

### Conclusion:
The set \( T(\text{Lip}_K[0, 1]) \) is equicontinuous.

Would you like further details or have any additional questions? Here are some related questions for further exploration:

1. What are the properties of bounded linear operators on Banach spaces?
2. How does the concept of equicontinuity relate to the Arzelà–Ascoli theorem?
3. What are the implications of equicontinuity for uniform convergence?
4. Can the concept of equicontinuity be extended to other function spaces?
5. How does the \( K \)-Lipschitz condition compare with Hölder continuity?

**Tip**: Understanding equicontinuity is crucial when dealing with families of functions in analysis, especially in the context of compactness and convergence theorems.

Explore the equicontinuity of bounded linear operators on Banach spaces with a focus on Lipschitz functions and their properties. Understand the conditions under which a mapping preserves the equicontinuity of a class of Lipschitz functions.

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