by Harmonic Sun
Last Updated August 14, 2019 09:20 AM - source

Say we have an integral transform $T$ that maps any function $f$ to the function $\varphi=T\left\lbrace f\right\rbrace$, defined by

$$\varphi(s)=\left[T\left\lbrace f\right\rbrace\right](s)=\int_a^bK(x,s)f(x)dx$$

With $K$ a given function of two variables, called the **kernel of the transform**.

The domain $A$ on which $T$ is defined is the set of all functions $f$ such that the integral is exists.

Suppose $T$ is **injective**, then it is **bijective** from $A$ to its range $B=T\left( A\right)$, and the **inverse transform** $T^{-1}:B\to A$, that recovers $f$ from $\varphi$, exists.

Now, for some integral transforms, I know that there is an analytical expression for $T^{-1}$ if $\varphi$ belongs to a certain subset of $B$:

$$f(x)=\left[T^{-1}\left\lbrace \varphi\right\rbrace\right](x)=\int_c^dK^{-1}(x,s)\varphi(s)ds$$

In which case, the function of two variables $K^{-1}$ is called the **inverse kernel** of $K$ (hence the $^{-1}$ exponent in the notation).

My question is, for the generic injective integral transform $T$ and its associated Kernel $K$ defined above, how can we tell that there is at most

onepossible inverse Kernel ? In other words, how can we tell that if there is an inverse kernel $K^{-1}$, it is necessarilyunique?

Naively, I can imagine a scenario where, although $T^{-1}$ is known to exist because of the injectivity of $T$, we don't have a analytical expression of $T^{-1}$ valid on all of its domain (i.e the range of $T$), but rather two integral transforms with each a different Kernel, defined on two disjoint subset of the domain $B$ of $T^{-1}$, such that $T^{-1}$ coincides with each of these on their respective domains...

Any insight ?

(if you think my question is ill-defined, feel free to suggest a more appropriate setting)

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