Discussion
Let \(X = \{f \in C[0,1] : f(0) = 0 = f(1)\}\) with the norm \(\|f\|_{\infty} = \sup_{0 \le t \le 1} |f(t)|\), where \(C[0,1]\) is the space of all real-valued continuous functions on [0, 1]. Let \(Y = C[0,1]\) with the norm \(\|f\|_{2} = \left(\int_{0}^{1} |f(t)|^{2} dt\right)^{1/2}\). Let \(U_{X}\) and \(U_{Y}\) be the closed unit balls in X and Y centred at the origin, respectively. Consider \(T: X \to \mathbb{R}\) and \(S: Y \to \mathbb{R}\) given by
\[ Tf = \int_{0}^{1} f(t)\,dt \]
and
\[ Sf = \int_{0}^{1} f(t)\,dt. \]
Consider the following statements:
S1: \(\sup_{f \in U_{X}} |Tf|\) is attained at a point of \(U_{X}\).
S2: \(\sup_{f \in U_{Y}} |Sf|\) is attained at a point of \(U_{Y}\).
Then, which one of the following is correct?
Answer: B
S2 is TRUE and S1 is FALSE.
S1: for f in \(U_{X}\), \(|Tf| \le \int_{0}^{1}|f| \le 1\), and functions equal to 1 except near the end points make \(Tf\) as close to 1 as we like, so the supremum is 1. But reaching 1 would need \(|f(t)| = 1\) everywhere, impossible for a continuous f with \(f(0) = f(1) = 0\). The supremum is not attained.
S2: by the Cauchy-Schwarz inequality, \(|Sf| = |\int_{0}^{1} f \cdot 1| \le \|f\|_{2}\|1\|_{2} \le 1\), with equality for the constant function f = 1, which lies in \(U_{Y}\) since its norm is 1. The supremum 1 is attained. So only S2 holds.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development