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I am interested in the convergence properties of $f_{n,m}$, with $n,m\in \mathbb N$, of the form $$ f_{n,m}=f(m/n) (1-e^{-n^2})+\delta_{m,0}\,n e^{-n^2} $$ in the limit $n\to\infty$. Here the function $f$ is analytic.

In principle, choosing $n,m\to\infty$, $m/n=x$ constant, we have that $$ \lim_{n\to\infty\\m/n=x}f_{n,m}=f(x), $$ where we can (I guess?) reconstruct the function $f(x)$ for all positive real $x$ for $f_{n,m}$.

My question concerns the case $x=0$, which can be obtained from the limit of $f_{n,0}$ or from $f_{n,m}$ at fixed $m$. Both will converge to $f(0)$ but not exactly in the same way. I'm not sure that speaking of uniform or non-uniform convergence makes sense here, as we are looking at the convergence of $f_{n,m}=f_n(x^{(m)}_n)$, where different choices of $x^{(m)}_n$ converge to the same $x$, but $f_n(x)$ is not continuous.

What can I say about the convergence of this series?

Adam
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  • You can say it doesn't converge... As you pointed out, it's easy to find subsequences converging to different values. See (https://math.stackexchange.com/questions/3204124/what-is-the-definition-of-double-sequence-a-mn-being-convergent-to-l) – 2by2is2mod2 Apr 23 '24 at 18:20
  • Thanks for the comment. In my case, this absence of a double limit is not a bug, but a feature (it corresponds to taking the continuum limit in a physics problem). What I'm really interested in is the different convergence to $f(x)$ for $x=0$ and $x\neq 0$. Due to the last term in the definition of $f_{n,m}$, we see that the convergence to $f(0)$ choosing $m=0$ or $m=100$ is not the same as the convergence to $f(1)$ choosing $m=n$ or $m=n+100$. – Adam Apr 24 '24 at 07:41
  • I'm not sure I understood your question. If you impose the condition $m/n = x$ and you study the limit for $n$ going to $+\infty$, then of course you get just a limit of a sequence of functions, which you can study as usual. You can speak about uniform convergence, once you give the domain of definition of $f$. Basically you can completely forget about $m$ and the Kronecker's delta just become a Dirac delta. – 2by2is2mod2 Apr 24 '24 at 17:30
  • You can impose different relations between $m$ and $n$ as well (like your $m = n + 100$) and then study that limit. But once you choose $m$ it is really just a usual limit of functions. – 2by2is2mod2 Apr 24 '24 at 17:42
  • Otherwise I'm really sorry I don't understand your question. Maybe you could try explain it a little more by saying what you mean with 'convergence to $f(0)$ choosing $m=0$ or $m=100$ is not the same as the convergence to $f(1)$ choosing $m=n$ or $m=n+100$'. – 2by2is2mod2 Apr 24 '24 at 17:47
  • @2by2is2mod2 To give some context, which might help: in physics, we often start by defining problems in finite size $L$. In Fourier space, we obtain functions in terms of "momenta" $p_m = 2\pi m/L$. We then take the limit $L\to\infty$, and then $p_m$ becomes a real variable, while $f_m$ becomes a function $f(p)$. Think of $n$ as the equivalent of $L$, and you have my problem: when I reconstruct my function $f(p)$ as $L\to\infty$, how to describe this convergence? What is the effect of this weird term at $m=0$? – Adam Apr 24 '24 at 21:24
  • I think continuum limit has not a rigorous mathematics explanation, but I could be wrong. Anyway, let's try to be rigorous and consider $m$ as a discrete parameter. Then the delta term is just 0 or 1 depending on $m$ being $0$ or not. That's all. In the continuum limit you will probably recognize it as a Dirac delta or something like that. But the problem is discrete (as it is) and you should really treat it like that. The continuum limit way of thinking here begins to show its limits. – 2by2is2mod2 Apr 24 '24 at 22:59
  • There could be some rigorous explanation of continuum limit (I don't know sincerely)... If it exists, there you'll likely find your answer! – 2by2is2mod2 Apr 24 '24 at 23:01

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