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Let $H=(H,(\cdot, \cdot))$ be a Hilbert space and $(u_n)_{n \in \mathbb{N}} \subset H$ a complete orthonormal set in $H$. Given $u \in H$, in some papers, for instance $[1, \text{Theorem 4.10}]$ and $[3, \text{page 257}]$, the authors consider a series of the form $$ \sum_{n\in \mathbb{N}\setminus J}\frac{1}{a_n}(u, v_n)v_n \cdot a_n \tag{1} $$ where $a_n \in \mathbb{R}$, for all $n \in \mathbb{N}$, and $J:=\{i \in \mathbb{N} \; ; \; a_i=0\}$ is a finite set. Obviously, by $(1)$ we have $$ \sum_{n \in \mathbb{N}\setminus J}\frac{1}{a_n}(u, v_n)v_n \cdot a_n=\sum_{n \in \mathbb{N}\setminus J}(u, v_n)v_n. $$

On the other hand, it is well know (see $[2$, Theorem $9.12$ $]$) that we can write $$ u=\sum_{n \in \mathbb{N}}(u, v_n)v_n = \sum_{n \in \mathbb{N}\setminus J}(u, v_n)v_n+\sum_{n \in J}(u, v_n)v_n. \tag{2} $$

Making a notational convention, namely $$ 0=\frac{0}{0} \tag{3} $$ the authors write then by $(1)$ and $(2)$ the following $$ \sum_{n\in \mathbb{N}\setminus J}\frac{1}{a_n}(u, v_n)v_n \cdot a_n =\sum_{n\in \mathbb{N}}\frac{1}{a_n}(u, v_n)v_n \cdot a_n = u. $$

Question. In this context, why the notational convention in $(3)$ can be considered?

In the book $[4, \text{Section $8.5$}]$ the division by zero is discussed, but I haven't been able to conclude whether or not I can even consider $(3)$. Any suggestion?

$[1]$ Albert, John P. Positivity Properties and Stability of Solitary–Wave Solutions of Model Equations For Long Waves. Communications in partial differential equations $17.1-2 (1992): 1-22.$

$[2]$ Bachman G. and Narici, L., Functional Analysis. New York: Academic Press, $2000$.

$[3]$ Freeden, Willi, and M. Zuhair Nashed. Ill-posed problems: operator methodologies of resolution and regularization. Handbook of Mathematical Geodesy. Birkhäuser, Cham, $2018. 201-314.$

$[4]$ Suppes, P., Introduction to logic. D. Van Nostrand Co., Inc., New York, $1957$.

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    $\begingroup$ I (probably naively) don't think it's worth dwelling on too much. It's a way to compactify the notation. Just as how $0^0$ is undefined, but we write $$p(x) = \sum_{i=0}^n a_i x^i$$ for polynomials, and still have them defined at $x=0$ by letting $0^0=1$. The alternative would be to write $$p(x) = a_0 + \sum_{i=1}^n a_i x^i$$ In this case we avoid having to get rid of $J$ each and every time to keep the expression well-defined. I don't know if there might be a deeper reason for this convention in this instance however. (Or if you want said deeper explanation and I'm overlooking it) $\endgroup$ Commented Aug 9, 2022 at 21:51
  • $\begingroup$ Except there is no real algebraic reason for $0^0$ to be undefined, while there is obvious reasons to treat $\frac00$ as undefined. But you are essentially correct, it's just notation, intended as a shorthand. As long as you know what it "really" means, and only use it in this context... @PrincessEev $\endgroup$ Commented Aug 9, 2022 at 21:56
  • $\begingroup$ @PrincessEev None of the above comments gave a satisfactory explanation, unfortunately. I would like a deeper and more mathematically coherent explanation. Since, we cannot to obtain the expression for $u$ if we don't able do sum over whole $\mathbb{N}$. $\endgroup$
    – Guilherme
    Commented Aug 9, 2022 at 22:48

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There is no serious mathematical content behind that convention... and certainly no assertion that $0/0=0$ in any numerical sense. Rather, to say this is a (rather cavalier) version of saying that, if $a_n=0$, then just drop that term entirely in the renormalized infinite sum.

The only small mathematical point is that the formula for properly renormalizing the coefficients is not correct for $a_n=0$... For other reasons, we know that in that case the $n$th renormalized coefficient should still be $0$...

I myself would indeed be disquieted by saying that a thing is true "by convention", as opposed to giving a good reason. In that regard, the author somewhat misrepresented the situation, but there's nothing serious hiding behind that glibness.

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  • $\begingroup$ So, is mathematically coherent to say that $0=\frac{0}{0}$? $\endgroup$
    – Guilherme
    Commented Aug 9, 2022 at 22:46
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    $\begingroup$ @Guilherme, not really. But, more to the point, we should not be finding ourselves in any situation where we actually have to do something with $0/0$. In the case at hand, the otherwise-correct formula for renormalization is not literally correct when $a_n=0$... and this "convention" does yield the correct renormalized coefficient... but we knew its correctness for other (simple!) reasons, not because of such craziness. $\endgroup$ Commented Aug 9, 2022 at 22:49
  • $\begingroup$ What you means by ''renormalization'' ? $\endgroup$
    – Guilherme
    Commented Aug 9, 2022 at 23:37
  • $\begingroup$ By "renormalization", I mean your second displayed (not-numbered) line. You multiplied-and-divided by $a_n$... which is dangerous if $a_n=0$, ... but the term would have been $0$ already if $a_n=0$, so, while the formulaic business that is correct for $a_n\not=0$ fails for $a_n=0$, it really doesn't matter, because those summands were $0$ from the outset... $\endgroup$ Commented Aug 9, 2022 at 23:46
  • $\begingroup$ Why the term would have been $0$ if $a_n=0$? That is, why $\frac{1}{a_n}(u, v_n)v_n \cdot a_n=0$ for each $n \in J$? As I understand it, that's what you mean.. $\endgroup$
    – Guilherme
    Commented Aug 9, 2022 at 23:53

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