Question

Let \( L^2[0, \pi] \) denote the space of all real valued Lebesgue square integrable functions on \( [0, \pi] \). Let \( T: L^2[0, \pi] \to L^2[0, \pi] \) be defined as follows: \[ T(f(x)) = \sin x \int_0^{\pi} f(t)\cos t \, dt + \cos x \int_0^{\pi} f(t)\sin t \, dt \]

Then the value of \( \dfrac{4}{\pi}\|T\| \) is equal to ______. (Answer in integer)

Hint:

T has rank 2, its whole range is spanned by sin x and cos x, which are orthogonal on [0, pi]; use that to bound the norm.

Answer 1

Correct Answer: 2

Step 1: Rewrite \( T \) using inner product notation.
Working in \( L^2[0,\pi] \) with inner product \( \langle f,g \rangle = \int_0^{\pi} f(t)g(t)\, dt \), the operator can be written as
\[ T(f) = \langle f, \cos t \rangle \sin x + \langle f, \sin t \rangle \cos x \]
So \( T \) sends every function \( f \) to some combination of just \( \sin x \) and \( \cos x \). This means \( T \) is a rank-\( 2 \) operator, its entire range is the \( 2 \)-dimensional space spanned by \( \sin x \) and \( \cos x \).

Step 2: Check that \( \sin x \) and \( \cos x \) are orthogonal on \( [0,\pi] \).
\[ \langle \cos t, \sin t \rangle = \int_0^{\pi} \cos t \sin t \, dt = \frac{1}{2}\int_0^{\pi} \sin 2t \, dt = \frac{1}{2}\left[-\frac{\cos 2t}{2}\right]_0^{\pi} = \frac{1}{4}(-1+1) = 0 \]
So \( \cos t \) and \( \sin t \) are orthogonal in \( L^2[0,\pi] \). Also their norms match:
\[ \|\cos t\|^2 = \int_0^{\pi}\cos^2 t\, dt = \frac{\pi}{2}, \qquad \|\sin t\|^2 = \int_0^{\pi}\sin^2 t\, dt = \frac{\pi}{2} \]
So \( \|\cos t\| = \|\sin t\| = \sqrt{\pi/2} \). Call this common value \( c \).

Step 3: Build an orthonormal pair and rewrite \( T \).
Let \( e_1 = \cos t / c \) and \( e_2 = \sin t / c \), unit vectors. Any \( f \in L^2[0,\pi] \) can be split as
\[ f = \alpha e_1 + \beta e_2 + g, \qquad \alpha = \langle f, e_1 \rangle, \ \beta = \langle f, e_2 \rangle \]
where \( g \) is orthogonal to both \( e_1 \) and \( e_2 \). Then \( \langle f, \cos t \rangle = c\alpha \) and \( \langle f, \sin t \rangle = c\beta \), so
\[ T(f) = c\alpha \sin x + c\beta \cos x = c\alpha (c \, e_2) + c\beta(c\, e_1) = c^2(\beta e_1 + \alpha e_2) \]

Step 4: Compute the operator norm.
Since \( e_1, e_2 \) are orthonormal, \( \|T(f)\|^2 = c^4(\alpha^2+\beta^2) \), while \( \|f\|^2 = \alpha^2+\beta^2+\|g\|^2 \ge \alpha^2+\beta^2 \). So
\[ \frac{\|T(f)\|}{\|f\|} \le \frac{c^2\sqrt{\alpha^2+\beta^2}}{\sqrt{\alpha^2+\beta^2}} = c^2 \]
with equality when \( g = 0 \), which is achievable by taking \( f \) as a pure combination of \( \cos t, \sin t \). So
\[ \|T\| = c^2 = \left(\sqrt{\frac{\pi}{2}}\right)^2 = \frac{\pi}{2} \]

Step 5: Compute the required value.
\[ \frac{4}{\pi}\|T\| = \frac{4}{\pi}\cdot\frac{\pi}{2} = 2 \]

Final Answer:
The value of \( \dfrac{4}{\pi}\|T\| \) is \( 2 \). \[ \boxed{2} \]