Let $I$ be the open interval $(0, 1)$ and $H := L^2 (I)$ equipped with the usual inner product. Consider the linear map $T: H \to H$ defined by $$ (Tf) (x) = \int_0^x t f(t) dt + x \int_x^1 f(t)dt, \quad x \in I. $$
I am trying to solve a problem in Brezis' Functional Analysis
Exercise 8.28
Check that $T$ is a bounded linear operator.
Check that $T$ is a compact operator.
In my below attempt, I use an auxiliary result from the same book. Is there another approach that does not need to appeal to this result?
We have $$ |(Tf) (x)| \le \left | \int_0^x t f(t) dt \right | + \left | x \int_x^1 f(t)dt \right |\le 2\int_0^1 |f(t)| dt \le 2\|f \|_{H}, $$ which implies $\| Tf \|_\infty \le 2\|f \|_{H}$. Clearly, $\| Tf \|_{H} \le \| Tf \|_\infty$.
Let $(f_n)$ be a bounded sequence in $H$. Because $H$ is reflexive, there is $f \in H$ and a subsequence (also denoted by $(f_n)$ for simplicity) such that $f_n \to f$ weakly. Then $$ \int_0^x t f_n(t) dt = \int_0^1 f_n(t) t 1_{(0, x)} (t) dt \xrightarrow{n \to \infty} \int_0^1 f (t) t 1_{(0, x)} (t) dt = \int_0^x t f (t) dt, $$ because the map $t \mapsto t 1_{(0, x)} (t)$ belongs to $H$. Similarly, we get $$ \int_x^1 f_n (t)dt \xrightarrow{n \to \infty} \int_x^1 f (t)dt. $$ It follows that $Tf_n \to Tf$ everywhere on $I$. Notice that $\| Tf_n \|_{L^3} \le \| Tf_n \|_\infty \le 2\|f_n \|_{H}$. The claim then follows from the following result (in the same book), i.e.,
Exercise 4.16.3 Let $(\Omega, \mathcal F, \mu)$ be a measure space with $\mu(\Omega) < \infty$. Let $p \in (1, \infty)$ and $f_n,f:\Omega \to \mathbb R$ be measurable such that $\sup_n \|f_n\|_p < \infty$ and that $f_n \to f$ $\mu$-a.e. Then $\|f_n - f\|_q \to 0$ for every $q \in [1, p)$.
