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Fourier transforms are famous as they turn convolution of two functions into multiplication.

Do there exist integral transforms which in a similar way turn function composition into multiplication?

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    $\begingroup$ An operation turned into multiplication by an integral transform must be commutative and linear in both factors. Function composition is neither. $\endgroup$ – Professor Vector Jul 10 '17 at 13:01
  • $\begingroup$ What if we allow to take them to a vector space somehow? Matrices don't need to be commutative. $\endgroup$ – mathreadler Jul 10 '17 at 13:54
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Lets start with the following assumptions on $\mathcal L$:

  1. $\mathcal L:D \to\Bbb R^\Bbb R$ is linear on its domain $D\subset \Bbb R^\Bbb R$. $D$ is a linear subspace of $\Bbb R^\Bbb R$.
  2. If $f,g \in D$ then $f\circ g\in D$ and $\mathcal L(f\circ g) = \mathcal L(f)\cdot \mathcal L(g)$.

The form of $\mathcal L$ is strongly limited by how big $D$ is. Consider the following statements:

Assume a constant function $c$ is in $D$, then: $$\mathcal L( c ) = \mathcal L(c \circ 0) = \mathcal L(c) \cdot \mathcal L(0)=0$$ so all constant functions must be sent to $0$. If one constant function is in $D$ then all are, further if the identity is in $D$ then translations by $x$, written $\tau_x= \mathrm{id}+c_x$, are in $D$ also and $$\mathcal L(f\circ \mathrm{id})=\mathcal L(f)\cdot \mathcal L(\mathrm{id})=\mathcal L(f), \quad \mathcal L(f\circ \tau_x) = \mathcal L(f)\cdot \mathcal L(\mathrm{id}+c_x)=\mathcal L(f)$$ so $\mathcal L(\mathrm{id})$ must be idempotent and multiplication with it must be identity on $\mathrm{im}(\mathcal L)$. Further the image of any function has to be translation invariant.

This implies that any compactly supported function with bounded image must be sent to zero as there must be $x$ so that $g\circ \tau_x \circ g$ is constant in this case. If you want $\mathcal L$ to have strong continuity properties, like for example if $g_\alpha\to g$ uniformly on compacta then $\mathcal L(g_\alpha)\to\mathcal L(g)$, this gives you that $\mathcal L$ must be zero, provided you have enough compactly supported functions in $D$.

So four "nice" conditions conflict to kill $\mathcal L$:

  1. The identity is in $D$.
  2. The constant functions are in $D$.
  3. There are enough compactly supported functions in $D$.
  4. $\mathcal L$ has relatively nice continuity properties.

There exist plenty of trivial non-zero $\mathcal L$ that do not have these four properties, for example if $D$ consists only of constant functions or of multiples of identity then the algebra with $\circ$ is homomorphic to $\Bbb R$.

Looking for a non-trivial non-zero $\mathcal L$ you should cross off a few of list elements. For example you could have $D= \mathrm{Aut}(\Bbb R)$, ie homeomorphisms $\Bbb R\to \Bbb R$ (this is closed under $\circ$). Another case could be $D=C_c(\Bbb R)$, that is compactly supported continuous functions. You can play around in similar ways to here to see some possibly interesting things, ie how can you model translations using only composition of elements in $C_c(\Bbb R)$?

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  • $\begingroup$ what is $\mathbb R ^\mathbb R$ $\endgroup$ – mathreadler Jul 10 '17 at 12:43
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    $\begingroup$ @mathreadler $\mathbb{R}^\mathbb{R}$ denotes the functions $\mathbb{R}\to\mathbb{R}$. More generally, $A^B$ are the functions $B\to A$. $\endgroup$ – haemi Jul 10 '17 at 13:06
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    $\begingroup$ $\Bbb R^\Bbb R = \{ f: \Bbb R\to \Bbb R\}$, here we can give this space the structure of an $\Bbb R$-algebra. I sort of ignore the question of what topology we want to give this space with the assumption that the more important $D$ will have some kind of a natural topology on it. $\endgroup$ – s.harp Jul 10 '17 at 13:11

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