Origin of the smooth but nowhere real analytic function built with dyadic rationals

I found the following interesting function and its analysis at Non-analytic smooth function article in Wikipedia. I include a screen capture below for those who don't wish to navigate away: I could not find a clear reference in the article as to the source of this example. Hence my question:

Question: who found the example above ?

I found the Fabius function article gives another example of a smooth but nowhere analytic function after reading this great MSE question but I need a reference for the nameless example I post above. Thanks in advance for your insight on this matter.

• There might be something relevant on p. 117 of du Bois-Reymond's 1883 paper that gave the first published example of a $C^{\infty}$ nowhere analytic function, but I can't read German so I'm just guessing. However, my feeling is that trying to track down your specific example might wind up being an ill-defined task, since it's probably contained in the many general examples (see this paper, for example) that have appeared over the years. The problem is here do you draw the line? – Dave L Renfro Jul 20 '18 at 12:10
• The problem is here do you draw the line? --- Of course, "here" should be "where". And what I meant is if someone has proved the $C^{\infty}$ and nowhere analyticity of a series whose terms are $a_n\cos(b_nx),$ where sequences $\{a_n\}$ and $\{b_n\}$ have certain general properties relative to some dense set, would this count? – Dave L Renfro Jul 20 '18 at 12:44
• @DaveLRenfro I would settle for the person who wrote the formula and conjectured, or partially proved, it is smooth but not real analytic. But, your educated opinion that this is common knowledge (if that is your opinion) is useful. We're using this function to build a counter-example so I just want to credit the source if appropriate and I'm not comfortable with a Wikipedia reference so... – James S. Cook Jul 20 '18 at 14:29
• I don't know whether this is a special case of a more general result (I definitely would NOT bet money against this being the case, however), let alone whether it's well known. It does have a kind of elegant simplicity about it, more so than the Mandelbrojt example. – Dave L Renfro Jul 20 '18 at 16:26
• @DaveLRenfro part of the simplicity of the example is that it is wrong! See math.stackexchange.com/q/2857965/36530. I have spent more time on this than I'd like to admit. – James S. Cook Jul 20 '18 at 20:57

Mandelbrojt (1942, p. 2) has $\displaystyle f(x)=\sum\limits_{n=1}^\infty e^{-\sqrt{n!}}\cos(n!x)$, with a very similar proof.