Past exam of the analysis course of the University of Oxford / 2008 / B / 8

Write $\lambda =a+ib$ with $b\ne 0$. Since $aI-T$ is self-adjoint, $⟨(aI-T)v,v⟩$ is real. The cross term in the squared norm therefore vanishes:This proves the equality and lower bound, and makes both $S=\lambda I-T$ and $S^{\ast }=\overline{\lambda }I-T$ injective.

If $S{v}_n$ converges, the lower bound shows $v_n$ is Cauchy. Completeness gives $v_n\to v$ and boundedness gives $S{v}_n\to Sv$, so the range is closed. Also $(\mathrm{ran}S{)}^{\perp }=\ker S^{\ast }=0$, so the range is dense. Being both closed and dense, it equals $V$. Hence $S$ is bijective, and its lower bound gives

Define $f^{♯}(s)=\overline{f(-s)}$. Interchanging the integrals and substituting variables givesMoreover Cauchy-Schwarz in the inner integral givesTaking the supremum over unit $g_2$ proves $\Vert T_f{g}_1{\Vert }_2\le \Vert f{\Vert }_1\Vert g_1{\Vert }_2$, so $T_f$ is bounded with norm at most $\Vert f{\Vert }_1$, and the adjoint identity is justified.

It remains to ensure necessity is a condition on the kernel itself. If $T_h=0$ for some $h\in L^1$, convolve $h$ with a smooth compactly supported approximate identity ${\eta }_{\epsilon }\in L^1\cap L^2$. Since $T_h{\eta }_{\epsilon }=0$, all these convolutions vanish. But $h\ast {\eta }_{\epsilon }\to h$ in $L^1$, by continuity of translations in $L^1$, so $h=0$ almost everywhere. Thus the map from kernels to convolution operators is injective. Applying this to $h=f-f^{♯}$ yields the exact criterionFor real-valued kernels this reduces to evenness; real-valuedness by itself is insufficient.