Question

Let \( 0<\alpha<1 \). Define \[ C^\alpha[0, 1] = \left\{ f : [0, 1] \to \mathbb{R} \ : \ \sup_{s \neq t, \, s,t \in [0, 1]} \frac{|f(t) - f(s)|}{|t - s|^\alpha}<\infty \right\}. \] It is given that \( C^\alpha[0, 1] \) is a Banach space with respect to the norm \( \| \cdot \|_\alpha \) given by \[ \| f \|_\alpha = |f(0)| + \sup_{s \neq t, \, s,t \in [0, 1]} \frac{|f(t) - f(s)|}{|t - s|^\alpha}. \] Let \( C[0, 1] \) be the space of all real-valued continuous functions on \( [0, 1] \) with the norm \( \| f \|_\infty = \sup_{0 \leq t \leq 1} |f(t)| \). 
If \( T: C^\alpha[0, 1] \to C[0, 1] \) is the map \( T f = f \), where \( f \in C^\alpha[0, 1] \), then which one of the following is/are TRUE?

• A. \( T \) is a compact linear map
• B. The image of \( T \) is closed in \( C[0, 1] \)
• C. The image of \( T \) is dense in \( C[0, 1] \)
• D. \( T \) is not a bounded linear map

Hint:

In functional analysis, a map between spaces with different norms may be compact, especially when the source space has stricter regularity conditions. Also, check if the image of the map is dense in the target space when the source space has more regularity.

Answer 1

The Correct Option is A, C

We are given that \( C^\alpha[0, 1] \) is a Banach space with respect to the norm \( \| \cdot \|_\alpha \), and we are analyzing the properties of the linear map \( T: C^\alpha[0, 1] \to C[0, 1] \) defined by \( T(f) = f \).

Step 1: Compactness of \( T \)
The map \( T \) is a compact linear map because it maps from the space of functions with regularity \( \alpha \) to a larger space of continuous functions. In such settings, maps that involve continuous embeddings from smaller, more regular spaces to larger spaces tend to be compact. Therefore, (A) is TRUE.

Step 2: Density of the image of \( T \)
The space \( C^\alpha[0, 1] \) consists of functions that are more regular than arbitrary continuous functions in \( C[0, 1] \). However, any continuous function on \( [0, 1] \) can be approximated arbitrarily closely (in the sup norm) by functions in \( C^\alpha[0, 1] \), as these functions are a subset of the polynomials. Thus, the image of \( T \) is dense in \( C[0, 1] \). Therefore, (C) is TRUE.

Step 3: Image of \( T \) and closedness in \( C[0, 1] \)
The image of \( T \) is not closed in \( C[0, 1] \), because the image consists of functions that are more regular (in terms of smoothness) than the arbitrary continuous functions in \( C[0, 1] \), making the image not closed. Thus, (B) is FALSE.

Step 4: Boundedness of \( T \)
Since \( T \) is a linear map and is defined on a Banach space, and it is a continuous embedding, \( T \) is indeed bounded. Therefore, (D) is FALSE.

Final Answer
\[ \boxed{(A)\ T\ \text{is a compact linear map}} \quad \text{and} \quad \boxed{(C)\ \text{The image of } T \text{ is dense in } C[0, 1]}. \]