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Not sure if this has been asked before. Recently, I've been thinking about differentiation and integration as linear operators acting on functional spaces. I know that integration cannot be properly described as the inverse of differentiation , as the latter is not injective. Still, it came to me that it may be possible to treat differentiation as a bijective operator, with integration as its inverse, if the domain is restricted to a particular quotient space that I will now outline. Consider the application $$\mathcal D:C^1(I) \to C^0(I), \qquad \mathcal D(f(x))=f'(x) \quad \forall x \in I,$$ where I have chosen the domain to be the space of differentiable real-valued functions with continuous derivative on the interval $I \subseteq \Bbb R$ for simplicity. Now to my question. Consider the equivalence relation $$r = \{(f,g) \in C^1(I) \times C^1(I) \ | \ \forall x \in I \quad f(x) = g(x)+k, \ k \in \Bbb R\}$$ and the resulting quotient space $C^1(I)/r$. In this new set, each function belongs to a unique equivalence class in which all elements differ by a constant. If we then restrict the domain of $\mathcal D$ to said quotient space, would the application become bijective, with integration (in the indefinite sense) as its inverse? And could this work for broader functional spaces as well?

Jose Avilez
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interspazia
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  • Yes. People do this, although rather informally. https://math.stackexchange.com/questions/3767159/how-should-one-understand-the-indefinite-integral-notation-int-fx-dx-in/3767192#3767192 – JonathanZ Feb 02 '24 at 19:13
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    For a fixed point $x_0\in I$ your quotient is isomorphis to $C^1_0(I)={f\in C^1(I):f(x_0)=0}$ and the inverse of $\mathcal D:C^1_0(I)\to C(I)$ is the unique anti-derivative which vanishes at $x_0$ (i.e., $\int_{x_0}^x g(t)dt$). Your wording that you restrict the domain of $\mathcal D$ to the quotient is not correct: You consider the unique factorization of $\mathcal D$ over the quotient. This priciple is much more general: Every linear operator $T:X\to Y$ induces a linear injection $X/\mathrm{kern}(T)\to Y$. – Jochen Feb 03 '24 at 07:44

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