For $T: L^2[0,1]\to L^2[0,1]$ compact operator given by $T(f(x))= \int_{0}^{x}f(t)dt$ do:
- Calculate $T^{*}$
- Calculate all eigenvalues and eigenvectors of $T^{*}T$
- $||T||=\frac{2}{\pi}$ My attempt was:
Firstly, i can try to evaluate $\langle Tf(x),g(x)\rangle = \int_{0}^{1}(\int_{0}^{x}f(t)dt)g(x)dx$ . If i do integration by parts, for example i choose $u=\int_{0}^{x}f(t)dt $ and $dv= g(x)dx$ implies that $du= f(x)$ and $v= \int_{0}^{1}g(x)dx$. Therefore,
$\langle Tf(x),g(x)\rangle= [\int_{0}^{x}f(t)dt \int_{0}^{1}g(x)dx]_{|_{0}^{1}}- \int_{0}^{1} (\int_{0}^{1}g(x)dx)f(y)dy = \int_{0}^{1}g(x)dx\int_{0}^{1}f(t)dt -\int_{0}^{1}f(y)dy\int_{0}^{1}g(x)dx$ looks as $0$, and the same time is equal to $\langle f(x),Tg(x)\rangle$ but i am not sure.
To do $2$ i need $1$ so i need help too.
For $3$ I see that $||T(f)||_{L^2}^2= \int_{0}^{1}|(\int_{0}^{x}f(t)dt)|^2dx \leq \int_{0}^{1}\int_{0}^{x}(|f(t)|dt)^2dx\leq \int_{0}^{1}\int_{0}^{1}(|f(t)|.1dt)^2dx\leq \int_{0}^{1}||f||_{L^2}dx$ then $||T||\leq 1$ but i thought that the functions $\sin nx$ and $\cos nx$ with $n\in \mathbb N$ give me the answer that the excersice tell me, i am right? or perhaps is something similar with the function $\sin$ and $\cos$ to appear the term $\frac{2}{\pi}$ in the norm of $T$. I will say thank you for the possible hints. Best
