Consider the following function $P$ defined as $P(f(x))=f'(x)-f(x)$

What is the inverse of $P$?

I can't figure this one out. A hint would be very helpful.

  • $\begingroup$ Did you look at some examples? $\endgroup$
    – Umberto P.
    Oct 13 at 22:05
  • $\begingroup$ What's the domain of $P$? $\endgroup$ Oct 13 at 22:06
  • 3
    $\begingroup$ So it is actually $P(f)=f'-f$ and we are looking for a functional $Q(P(f))=P(Q(f))=f$? $\endgroup$ Oct 13 at 22:16
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    $\begingroup$ P is not one-to-one. For instance, both e^x and 2e^x are mapped to zero. Is there any more information you’re leaving out? $\endgroup$
    – Rob Dukes
    Oct 13 at 22:18
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    $\begingroup$ @John.W $P$ and $Q$ are not invertible in these cases either, since their kernels are still nontrivial. Notice that $P\left(ce^{ax}\right)\equiv 0$ and $Q\left(ce^{-bx}\right)\equiv 0$ for any constant $c$. $\endgroup$ Oct 13 at 22:26

While there is no inverse, you can find a right inverse. As Rezha Adrian Tanuharja points out, we have $$\tag{$1$} P(f(x)) = e^x\frac{d}{dx}\left(e^{-x}f(x)\right). $$ But then we can solve for $f$ to get $$\tag{$2$} f(x) = e^x \int e^{-x}P(f(x))\,dx, $$ for some choice of antiderivative. More importantly, $(2)\Rightarrow(1)$ for any choice of antiderivative, which means that $$ Q(g):= x \mapsto e^x \int_a^x e^{-t}g(t)\,dt $$ satisfies $P(Q(g))=g$. Note that $Q$ is not a unique solution, since any antiderivative would have worked.

As many have noted, $P$ is not injective, so you can't find a well defined left inverse.


There is no inverse. The function is not injective. $\text P(0) = \text P(e^x)$.

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    $\begingroup$ Maybe you can add more to the answer? $\endgroup$ Oct 13 at 23:10
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    $\begingroup$ Where is the proof ? $\endgroup$
    – jimjim
    Oct 14 at 1:35
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    $\begingroup$ @jimjim P(0) = P(e^x) $\endgroup$ Oct 14 at 2:20
  • $\begingroup$ Thank you, please add a few more comments for others as well to understand why that is so and why it implies that the function is not injective $\endgroup$
    – jimjim
    Oct 14 at 2:50

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