How do I prove the solution operator for my elliptic PDE is bounded?

Question

For all $f\in L^2(0,1)$, I know tat there exists a unique $T(f) \in H^1(0,1)$, such that $$-T(f)''(x)+xT(f)'(x)+T(f)(x)=f(x) \quad \forall x\in(0,1).$$

I am wondering how to prove that $T(f)$ is bounded? Could anyone give me a hint?

Answer 1

Huge hint. From your previous question PDE Sobolev Space, prove the equation holds weakly

there exists a unique weak solution $u=T(f)$ in $H^1_0(0,1)$ in the sense that $$B[u,v] = \int_0^1 fv \,dx$$ for all $v\in H^1_0$, where $B$ is the associated bilinear form. $B$ was proven coercive $$B[u,u]\ge \theta \|u\|_{H^1_0}^2$$ for some $\theta>0$. These two things combined prove that the map $f\mapsto T(f)$ is bounded as a map $L^2\to H^1_0$ with operator norm $\frac1\theta$.