Necessary condition for invertible integral tansform.

Question

i have a rather complicated question (at least for me) to figure out. Let’s say that: $$I[f(t)](u)=\int_{t_0}^{t_1}f(t)K(t,u) \mathrm{d}t$$ Is a general intengral transform $I$ with kernel $K$. Are there any necessary and/or sufficient conditions that $K$ should satisfy in order to make $I$ into an invertible integral transform? Alternatively, when is $I$ invertible?

Answer 1

I take the liberty of changing a bit your notation. Let's denote a (linear) integral transform and its inverse tranform by $$ \left\{ \begin{array}{ccccl} F(\omega) &:=& I[f(t)](\omega) &=& \displaystyle \int_a^b f(t)\,k(t,\omega) \,\mathrm{d}t \\ f(t) &:=& I^{-1}[F(\omega)](t) &=& \displaystyle\int_\alpha^\beta F(\omega)\,K(\omega,t) \,\mathrm{d}\omega \end{array} \right. $$ where $K$ is the "inverse" kernel of $k$. The condition of invertibility translates as $I^{-1} \circ I = I \circ I^{-1} = \mathrm{id}$, hence the following relations : $$ \left\{ \begin{array}{ccccl} f(t) &=& \displaystyle \int_\alpha^\beta F(\omega)\,K(\omega,t) \,\mathrm{d}\omega &=& \displaystyle \int_\alpha^\beta\mathrm{d}\omega \int_a^b\mathrm{d}t' \,f(t')k(t',\omega)K(\omega,t) \\ F(\omega) &=& \displaystyle \int_a^b f(t)\,k(t,\omega) \,\mathrm{d}t &=& \displaystyle \int_a^b\mathrm{d}t \int_\alpha^\beta\mathrm{d}\omega' \,F(\omega')k(t,\omega)K(\omega',t) \end{array} \right. $$ Finally, we deduce from these relations the following conditions on the kernels : $$ \left\{ \begin{array}{ccl} \displaystyle \int_\alpha^\beta \,k(t,\omega)K(\omega,t') \,\mathrm{d}\omega &=& \delta(t-t') \\ \displaystyle \int_a^b \,K(\omega,t)k(t,\omega') \,\mathrm{d}t &=& \delta(\omega-\omega') \end{array} \right. $$ with $\delta$ the Dirac delta function.