Let α be positive and define linear operator T from L2[0,1] to L2[0,1] by Tf(x)=(∮[x→1]f(t)dt)-αxf(x).Then,T is bounded. Find the range of α such that R(T) is dense in L2[0,1].
I understand T is bounded linear operator on L2[0,1]. So,I want to use the following equation:cl(R(T))=(kerT*)⊥ Then,R(T) is dense in L2[0,1] if and only if kerT*={0} However,I don’t know how to find the adjoint operator of T. Please give me a hint or answer.
Assuming that you wanted to write $Tf:=\int f\,\mathrm d \lambda -\alpha \operatorname{id}f$ for some $\alpha >0$ then you can set $Rf:=\int f\,\mathrm d \lambda $ and $S:=\alpha \operatorname{id}f$, then note that $T^*=R^*-S^*$ and $$ \langle Rf,g \rangle=\int \left(\int f\,\mathrm d \lambda \right)\bar g\,\mathrm d \lambda =\int \overline{\left(\int g\,\mathrm d \lambda \right)}f\,\mathrm d \lambda =\langle f,R^*g\rangle\\ \therefore \quad R^*g=Rg=\int g\,\mathrm d \lambda $$ Similarly we find that $S^*g=\overline{\alpha \operatorname{id}}g$.
In the case that $Rf:=\int_x^1 f(t)\,\mathrm d t$ then we have that $$ \begin{align*} \langle Rf,g \rangle&=\int_0^1\left(\int_x^1 f(t)\,\mathrm d t\right)\bar g(x) \,\mathrm d x\\&=\int_{\{(x,t):0\leqslant x\leqslant t\leqslant 1\}}f(t)\bar g(x)\,\mathrm d (x,t)\\ &=\int_0^1\left(\int_0^t\bar g(x)\,\mathrm d x\right)f(t)\,\mathrm d t\\ &=\int_0^1\overline{\left(\int_0^t g(x)\,\mathrm d x\right)}f(t)\,\mathrm d t\\ &=\langle f,R^*g \rangle \end{align*}\\ \therefore \qquad R^*g=\int_0^t g(x)\,\mathrm d x $$