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I am wondering about whether there is a default or standard interpretation of statements such as $$\sum_{n=1}^\infty f_n(x) = f(x)$$ or equivalently

$$\sum_{n=1}^\infty f_n = f$$

In some cases these statements can mean 'uniformly convergent to $f$' or just 'pointwise convergent to $f$'. But sometimes I come across these equalities without the uniform or pointwise qualification, and thus in these situations I don't know whether as a default to interpret them as meaning pointwise or uniform convergence.

For example, when I first learnt about power series, we had not yet met the notions of uniform convergence (or pointwise). We simply defined $f(x) = \sum_{n=1}^\infty a_nx^n$. In hindsight, this equality really is equivalent to asserting the pointwise convergence of the series to $f$ over the radius of convergence. (Although it also turns out to be uniformly convergent within the radius)

Another example comes from the second answer in this question: When can a sum and integral be interchanged?, from the user Jonas Teuwen. In particular, he states that $f = \sum_n f_n$ in his answer. How should these equalities be interpreted? Is there a default, e.g. just assume it means pointwise, or is it entirely context dependent?

[Note: my current understanding is that when we deal with infinite series of functions, writing it as an equality is really a shorthand for some first order logic statement. I.e. it is completely analogous to the fact that stating $\lim_{n \rightarrow \infty} a_n = l$ in the case of real sequences really means $\forall \epsilon >0 \exists N \forall n>N (|a_n - l|< \epsilon)$. In this way I think of the equality symbol as just shorthand for a more verbose expression when it comes to series of functions, rather than meaning equality of mathematical objects so to say. In this sense, I don't know how to interpret the statements abut equality of series without any context.]

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    $\begingroup$ AFAIK there is no convention, it depends strongly in the context. It can represent pointwise convergence, or norm convergence, or almost everywhere convergence, or something else. In the link, as every $f_n\ge 0$ and the context is Lebesgue integration, then the convergence is pointwise $\endgroup$
    – Masacroso
    Jul 27 '20 at 14:46
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    $\begingroup$ I always take it to mean pointwise convergence. In the question you linked, I'm sure pointwise convergence is intended. $\endgroup$
    – saulspatz
    Jul 27 '20 at 14:46
  • $\begingroup$ "Although it also turns out to be uniformly convergent within the radius" That's not true. For example $\sum x^n$ does not converge uniformly on $(-1,1).$ $\endgroup$
    – zhw.
    Jul 27 '20 at 15:34
  • $\begingroup$ @zhw. Isn't it uniformly convergent on any subintervalof $(-1,1)$? I think my wording is unclear either way.. $\endgroup$
    – masiewpao
    Jul 27 '20 at 15:47
  • $\begingroup$ It's uniformly convergent on any $[-1+\delta,1-\delta].$ $\endgroup$
    – zhw.
    Jul 27 '20 at 16:03
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By default, it means that the convergence is pointwise. If it's meant to be uniform convergence, then it's either mentioned, or you are working in a function space with a topology where convergence means uniform convergence.

Like the normed space $(C^0(K),\Vert\cdot\Vert_\infty)$ of continuous functions $K\longrightarrow\mathbb R$ on a compact set $K$ with the norm $\Vert f\Vert_\infty:=\sup_{x\in K}\vert f(x)\vert$. Convergence in this normed space is equivalent to uniform convergence.

Or you could consider the topological space $(\mathcal H(\mathbb C),\tau)$ of holomorphic functions on $\mathbb C$ with the topology of compact convergence $\tau$. There, $\sum_n f_n=f$ would usually be interpreted to mean compact convergence.

So essentially, if it's explicitly mentioned that you're working in a function space whose topology implies a specific kind of convergence, then that kind of convergence is probably implied. Otherwise, the default is pointwise.

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