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Suppose $f$ is a continuous function of infinitely many real variables, and that 0 is an "identity element" for $f$ in the sense that

$$ f(0,\alpha,\beta,\gamma,\dots) = f(\alpha,\beta,\gamma,\dots). $$

Has anyone thought about the following limit (in particular, is there anything in the literature on it)?

$$ \lim_{\Delta\alpha\to0}\frac{f(\Delta\alpha,\ \alpha,\ \beta,\ \gamma,\dots) - f(\alpha+\Delta\alpha,\ \beta,\ \gamma,\ \dots) }{\Delta\alpha} $$

In case it makes anyone feel any better, for my purposes it may suffice to assume all but finitely many of the variable are zero (but of course with no prior finite bound on how many nonzero ones there are).

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  • $\begingroup$ Could you elaborate on the setting you are interested in? Is there a reason you don't just take the usual partial derivative of $f$ with respect to $\alpha$ and keep the other variables fixed? Do you have an example $f$ in mind? It is probably important to pin down whether you are considering only finitely many nonzero variables or not and then to put a topology on the domain of $f$ for this to make any sense. $\endgroup$
    – Noah Stein
    Commented May 25, 2010 at 19:54
  • $\begingroup$ Why would anyone want to consider such a limit? $\endgroup$
    – George Lowther
    Commented May 25, 2010 at 19:55
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    $\begingroup$ Being a bit naive here: isn't the limit you wrote equal to $ \lim \frac{f(\triangle \alpha,\alpha,\beta,\ldots) - f(0,\alpha,\beta,\ldots)}{\triangle \alpha} - \frac{f(\alpha + \triangle\alpha,\beta,\ldots) - f(\alpha,\beta,\ldots)}{\triangle \alpha} $ using that $f$ is right-translation invariant? Which formally makes it just the difference of the two derivatives. Also, what is the topology you are using on "$\mathbb{R}^\infty$"? I.e. what do you mean by continuous? Just separately continuous in each of the variables? $\endgroup$
    – Willie Wong
    Commented May 25, 2010 at 20:13
  • $\begingroup$ Applied to the function $$ g_k(\theta_1,\theta_2,\theta_3,\dots) = \sum_{|A|=k}\prod_{i\in A}\sin\theta_i\prod_{i\not\in A}\cos\theta_i $$ this seems to yield (but correct me if I'm wrong) $$ -(\cos\theta_1) g_{k-1}(\theta_2,\theta_3,\dots) - (\sin\theta_1)g_k(\theta_2,\theta_3,\dots) - \frac{\partial}{\partial\theta_1} g_k(\theta_1,\theta_2,\theta_3,\dots). $$ $\endgroup$
    – Michael Hardy
    Commented May 25, 2010 at 21:44

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I think Willie Wong's answer here suffices. This is just a case of looking at something from a suddenly different point of view and missing the obvious since it's not the way I'd been looking at it before.

I had an occasion to think about the difference $$ f(\alpha,\beta,\gamma,\delta,\dots) - f(\alpha+\beta,\gamma,\delta,\dots) $$ and was just playing around with that.

Continuity in the first variable is enough for the present purpose; I mentioned it only to rule out an obvious reason why the limit might not exist.

The functions I was thinking about were somewhat similar to the sum of products I mentioned in one of my comments above.

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  • $\begingroup$ @Michael Hardy: please accept this (your own) answer as an answer? This will prevent the MathOverflow Robot from automatically bumping this question up to the home page. $\endgroup$
    – Willie Wong
    Commented Nov 18, 2010 at 11:44

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