Continuity in infinite product

Question

if I define

$$f(x)=\begin{cases} \prod_{i=1}^{\infty} (e^x-e^{a_i})\cdot(e^x-e^{b_i})& \text{if }\ x\in [a_i,b_i] \\0 & \text{otherwise}\end{cases}$$ then what can we say about continuity of $f$.

Answer 1

I don't know if there are explicit theorems about the possibility of interchanging limit and infinite product, but we can exploit the uniform convergence theorem for sums:

Theorem

Let $X\subseteq \mathbb{R}$, $a\in\mathbb{R}\cup\{\infty\}$ be a limit point of $X$, $f_k:X\to \mathbb{R}$. If the limit $\lim_{x\to a}f_k(x)$ exists and $\sum_{k=0}^{\infty}f_k$ converges uniformly then \begin{equation}\lim_{x\to a}\sum_{k=0}^{\infty}f_k(x)=\sum_{k=0}^{\infty}\lim_{x\to a}f_k(x)\end{equation}

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Let me rewrite our function $f$ in the following equivalent form: $$ f(x)=\exp({\log f(x)})=\exp\left(\sum_{i=1}^\infty(\log (e^x-e^{a_i})+\log(e^x-e^{b_i}) )\right) $$ Let

$$ f_i(x):=\log (e^x-e^{a_i})+\log(e^x-e^{b_i}) $$ Both the limits $\lim_{x\rightarrow a_i}f_i(x)=-\infty$ and $\lim_{x\rightarrow b_i}f_i(x)=-\infty$ exist and $\sum_{i=1}^\infty f_i(x)$ converges uniformly (otherwise $f$ would not be well-defined). Hence $$ \begin{split} \lim_{x\rightarrow a_i}\exp({\log f(x)})&=\exp\left(\lim_{x\rightarrow a_i}\log f(x)\right)=\exp\left(\lim_{x\rightarrow a_i}\sum_{i=1}^\infty(\log (e^x-e^{a_i})+\log(e^x-e^{b_i}))\right)\\ &=\exp\left(\sum_{i=1}^\infty\lim_{x\rightarrow a_i}(\log (e^x-e^{a_i})+\log(e^x-e^{b_i}))\right)=[e^{-\infty}]=0, \end{split} $$ where I have applied the theorem above in the third equality and proceed similarly to obtain $\lim_{x\rightarrow b_i}\exp(\log f(x))=0$ (the expression $[e^{-\infty}]$ is just a notation for heuristic interpretation of the limit, just to show you why the limit is $0$).

You proceed exactly in the same way to show that $\lim_{x\rightarrow b_i}f(x)=0$. This shows that $f$ is continuous at $a_i$ and $b_i$ (and it is obviously continuous at any other point).