Let \( C[0, 1] \) be the Banach space of real valued continuous functions on [0, 1] equipped with the supremum norm. Define \( T: C[0, 1] \to C[0, 1] \) by
\[
(Tf)(x) = \int_0^x t f(t) \, dt.
\]
Let \( R(T) \) denote the range space of \( T \). Consider the following statements:
\[\begin{array}{rl} \bullet & \text{P: \( T \) is a bounded linear operator.} \\ \bullet & \text{Q: \( T^{-1}: R(T) \to C[0, 1] \) exists and is bounded.} \\ \end{array}\]
Show Hint
- both P and Q are TRUE
- P is TRUE and Q is FALSE
- P is FALSE and Q is TRUE
- both P and Q are FALSE
The Correct Option is B
Solution and Explanation
Step 1: Bounded Linear Operator.
The operator \( T \) is defined by an integral. It is linear because integration is a linear operation, and it is bounded because the supremum norm is preserved under integration.
Step 2: Inverse of \( T \).
For \( T^{-1} \) to exist, the range space \( R(T) \) must be bijective, meaning every element in \( R(T) \) should correspond to exactly one element in \( C[0, 1] \). However, \( T \) is not invertible because \( T \) is not surjective—there are continuous functions in \( C[0, 1] \) that cannot be obtained by applying \( T \) to any function in \( C[0, 1] \). Thus, \( T^{-1} \) does not exist.
Step 3: Conclusion.
Therefore, the correct answer is (B): \( T \) is a bounded linear operator, but \( T^{-1} \) does not exist.
Final Answer: \[ \boxed{(B) \text{P is TRUE and Q is FALSE}} \]
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