= Convolution kernel recovered from its L2 operator
{title2=$T_h=0\text{ on }L^2(\mathbb R),\ h\in L^1\implies h=0$}
An integrable <kernel> is uniquely determined by its <convolution> operator on $L^2$. Convolve it with a smooth compactly supported <approximate identity> $\eta_\varepsilon$, whose members also lie in $L^2$. A zero operator makes $h*\eta_\varepsilon=0$, while <translation continuity in Lp> gives $h*\eta_\varepsilon\to h$ in $L^1$, proving the assertion. Since the <adjoint operator> of <convolution> by $f$ is <convolution> by $f^\sharp(s)=\overline{f(-s)}$, this also proves the exact self-adjointness criterion $f=f^\sharp$ almost everywhere, including necessity at the level of kernels.
Back to article page
