Consider the operator $T:L^2(0,1)\rightarrow L^2(0,1)$. Is it well defined, linear, bounded, compact?

Question

I'm studying Funtional Analysis and I'm doing some exercise, but I haven't got any idea how to start this.

Consider the operator $T:L^2(0,1)\rightarrow L^2(0,1)$ given by $$(Tu)(x):=\int_0^1(x^2+3x+1)u(y)\ dy$$

Is it well defined, linear, bounded, compact?

My first idea is to get out the $p(x)$ from the integral being in $dy$, so it becomes

$$(x^2+3x+1)\int_0^1u(y)\ dy$$

but I don't think I can do something like that, right? Someone could help me?

Answer 1

$\int_{0}^{1}|Tu(x)|^2 = \int_{0}^{1}(x^2 +3x+1)^2 dx (\int_{0}^{1}u)^2dy\leq C ||u||^2$ by Jensen's inequality . Therefore $T$ is well-defined and bounded.For compactness let $||u_n||\leq C$, now as $L^2$ is reflexive $\{u_n\}$ has a weakly convergent subsequence $\{u_{n_{k}}\}$ i.e $\int_{0}^{1} u_{n_{k}} \rightarrow \int_{0}^{1} u $.So $Tu_{n_{k}}$ is convergent in $L^2(0,1)$. Thus compact.