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I am looking for the answer of the following derivative $$ \frac{\partial }{\partial f}\int_{\Omega} f(x)dx $$ where $\Omega$ is an open domain in $\mathbb{R}^n$ and $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is an integrable function. I feel like it should be something like $1$.... but I am not sure if it is true or not. We can do something for it by the calculus of variation.. For example, set $$I[f+\varepsilon \eta]=\int_\Omega [f+\varepsilon\eta](x)dx$$ and $$\frac{\partial}{\partial\varepsilon} I[f+\varepsilon \eta]\bigg|_{\varepsilon=0}=\int_\Omega \eta(x)dx$$ Of course, in this case, if I set $\eta(x)=\frac{1}{|\Omega|}$, I can get the answer I expected... But... anyway we have $\eta$ at the end which implies that the derivative is not well defined...

Is there any way to well define such partial derivative?

Lev Bahn
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  • I see answer accepted, but let me suggest for exact approach think about https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative – zkutch May 09 '21 at 23:39

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What does it mean to rigorously take the partial derivative with respect to a function? If you want only a formal/ "physics" argument, you could say something like \begin{equation} \frac{\partial}{\partial f}\int_\Omega f(x)dx=\int_\Omega \frac{\partial}{\partial f}f(x)dx=\int_\Omega dx=|\Omega|, \end{equation} where taking the derivative within the integral is justified for $f$ a nice enough function.

Andrew McMillan
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  • Thank... Actually, it is what I want. I am assuming $f$ is compactly supported smooth function so it is indeed nice function. Honestly I have not seen such direct computation... Is it really allowed if $f$ is nice enough? Can I ask a reference? – Lev Bahn May 09 '21 at 16:52
  • Since taking the partial derivative is just taking a limit, one can use the Dominated Convergence Theorem to justify commuting the integral with the limit. – Andrew McMillan May 09 '21 at 16:55
  • Hmm.... Yeah.. you are right... That was too easier than I thought... Then it seems like our $f$ does not have to have nice regularity as $$\frac{f+h-f}{h}=1$$ for any function. Can I ask your opinion? – Lev Bahn May 09 '21 at 16:59
  • Yes, I agree. I mean the integral has to be well-defined, so in that sense $f$ is nice as in its integrable. – Andrew McMillan May 09 '21 at 17:09