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)
Show 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.
Step 1: Express $T(f)$ in terms of two numbers.
Write $a = \int_0^{\pi} f(t)\cos t\, dt$ and $b = \int_0^{\pi} f(t)\sin t\, dt$, so $T(f) = a\sin x + b\cos x$. The operator norm $\|T\|$ is the smallest number $M$ such that $\|T(f)\| \le M\|f\|$ for every $f$, with equality achieved for some $f$.
Step 2: Compute $\|T(f)\|^2$ using orthogonality of $\sin x$ and $\cos x$ on $[0,\pi]$.
$\int_0^{\pi}\sin t\cos t\, dt = 0$, and $\int_0^{\pi}\sin^2 t\, dt = \int_0^{\pi}\cos^2 t\, dt = \frac{\pi}{2}$. So
\[ \|T(f)\|^2 = a^2\int_0^{\pi}\sin^2 x\, dx + b^2\int_0^{\pi}\cos^2 x\, dx = \frac{\pi}{2}(a^2+b^2) \]
Step 3: Bound $a^2+b^2$ using the Cauchy-Schwarz inequality.
By Cauchy-Schwarz, $a^2 = \langle f,\cos t\rangle^2 \le \|f\|^2\|\cos t\|^2 = \|f\|^2 \cdot \frac{\pi}{2}$, and similarly $b^2 \le \|f\|^2\cdot\frac{\pi}{2}$. Since $a$ and $b$ come from projecting $f$ onto two orthogonal directions, writing the normalized components of $f$ along $\cos t$ and $\sin t$ as $x_1, x_2$ gives $a^2+b^2 = \frac{\pi}{2}(x_1^2+x_2^2) \le \frac{\pi}{2}\|f\|^2$, because $x_1^2+x_2^2$ never exceeds the squared norm of $f$ itself.
Step 4: Combine the two bounds.
\[ \|T(f)\|^2 = \frac{\pi}{2}(a^2+b^2) \le \frac{\pi}{2}\cdot\frac{\pi}{2}\|f\|^2 = \left(\frac{\pi}{2}\right)^2\|f\|^2 \]
Taking square roots gives $\|T(f)\| \le \frac{\pi}{2}\|f\|$. This bound is reached exactly when $f$ is already a pure combination of $\cos t$ and $\sin t$ with nothing left over, so it is the sharp value of the norm:
\[ \|T\| = \frac{\pi}{2} \]
