How can I show that $T : L^2(0, \pi/2) \to L^2(0, \pi/2)$ defined as $T(f) = x + \frac{1}{4}\cos(x)\int_{0}^{\frac{\pi}{2}}f(y)dy$ is a contraction?

Question

I've tried the following:

\begin{equation} \|T(f)-T(g)\| = \|\frac{1}{4}cos(x)\int_0^{\frac{\pi}{2}}f(y) - g(y)dy\| = \frac{1}{4}|\int_0^{\frac{\pi}{2}}f(y) - g(y)dy| \|cos(x)\| = \frac{\sqrt{\pi}}{8}|\int_0^{\frac{\pi}{2}}f(y) - g(y)dy| \end{equation}

How can I relate this with $\|f-g\| = \sqrt{\int_0^{\frac{\pi}{2}}(f(y) - g(y))^2dy}$ ?

Answer 1

Let $f,g\in L^2(0,\pi/2)$. Then \begin{align*} \|Tf-Tg\|_{L^2(0,\pi/2)}^2&= \frac 1{16} \int_0^{\pi/2}\cos^2(x) \bigg ( \int_0^{\pi/2} f(y)-g(y) \, dy \bigg )^2 \, dx\\ &=\frac 1{16}\bigg ( \int_0^{\pi/2} f(y)-g(y) \, dy \bigg )^2 \int_0^{\pi/2}\cos^2(x) \, dx \\ &= \frac \pi{64}\bigg ( \int_0^{\pi/2} f(y)-g(y) \, dy \bigg )^2 \end{align*} By Hölder's inequality, \begin{align*} \bigg \vert \int_0^{\pi/2} f(y)-g(y) \, dy \bigg \vert &\leqslant \int_0^{\pi/2} \vert f(y)-g(y) \vert \, dy \\ &\leqslant \bigg ( \int_0^{\pi/2} 1 \, dy \bigg )^{1/2} \bigg ( \int_0^{\pi/2} \vert f(y)-g(y) \vert^2 \, dy \bigg )^{1/2} \\ &= (\pi/2)^{1/2} \bigg ( \int_0^{\pi/2} \vert f(y)-g(y) \vert^2 \, dy \bigg )^{1/2}. \end{align*} Hence, \begin{align*} \|Tf-Tg\|_{L^2(0,\pi/2)}^2 &\leqslant \frac {\pi^2}{128}\int_0^{\pi/2} \vert f(y)-g(y) \vert^2 \, dy = \frac {\pi^2}{128} \| f-g\|_{L^2(0,\pi/2)}^2. \end{align*} Thus, $$ \|Tf-Tg\|_{L^2(0,\pi/2)}\leqslant \frac{\pi}{4\sqrt2}\| f-g\|_{L^2(0,\pi/2)}$$ and $\frac{\pi}{4\sqrt2} \approx 0.56<1$.