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Let $f(x) , g_1(x) , g_2(x) , \ldots$ be analytic on $[-1,1].$

If for almost Every $x $e $[-1,1] $we have :

Property A :

If $$f(x) = g_1(x) + g_2(x) + \cdots \tag{property I }$$

And If $$f ' (x) = g_1 ' (x) + g_2 ' (x) + \cdots \tag{property II}$$

Then

$$f '' (x) = g_1 '' (x) + g_2 '' (x) + \cdots$$

And by induction the $n $th derivative satisfies ( $n$ is a positive integer )

$$f^{(n)} (x) = g_1^{(n)} (x) + g_2^{(n)} (x) + \cdots$$


notice property I does not always imply property II ;

example

The Fourier series for $f(x) = x.$


How to prove or disprove this ?

What are nice (counter-)examples ?


I got this question from the tetration forum where my mentor ( Tommy1729 ) posted it :

http://math.eretrandre.org/tetrationforum/showthread.php?tid=1088

I think i have seen it on sci.math as well ( also posted by him) I assume he knows the answer , but wants to see how people handle it.


I was reading complex analysis before I read his post. That inspired me to conjecture a variant : ( that I believe is easier )

Let $f(z),g_1(z),g_2(z),...$ be analytic on the closed unit circle.

The rest is equivalent.


I think replacing analytic with $C^\infty$ in Tommy's statement is interesting. Afterall there are functions that are $C^\infty$ almost everywhere but analytic almost nowhere.


This problem seemed simple when i first saw it , but is appears deceptively complicated to me. Or Maybe I am weak.

I considered fourier series , Riemann's series theorem and contour integrals to find a counterexample , but I failed.

I think it is possible to show the hypothesis correct if all the $f,g_i$ are monotone.

What bothers me most is - unlike most statements - that I am not even sure if it likely true or likely false !?

A further complication is summability methods, but for now Lets assume we consider ordinary sums that converge.

Is this a hard question ? Is this a new question ?

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  • $\begingroup$ Should i post on MO too ? $\endgroup$ – mick Aug 19 '16 at 20:58
  • $\begingroup$ I would use $g_1(x)$ rather than $g1(x)$, which is given by \$g_1(x)\$ $\endgroup$ – Simply Beautiful Art Aug 19 '16 at 23:16
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Consider this example:

$$\cos x = (\cos x - \cos (2x)/2^2) + (\cos (2x)/2^2 - \cos (3x)/3^2)+(\cos (3x)/3^2 - \cos (4x)/4^2) + \cdots.$$

Differentiating, we wonder if

$$-\sin x = (\cos x - \cos (2x)/2^2)'+ (\cos (2x)/2^2 - \cos (3x)/3^2)' + \cdots$$ $$ = (-\sin x + \sin (2x)/2) + (-\sin (2x)/2 + \sin (3x)/3) + \cdots.$$

Yes, this is fine: The $n$th partial sum is $-\sin x + [\sin(n+1)x]/(n+1) \to -\sin x.$ But we run into trouble with the next derivative, because the series of second derivatives is

$$(-\cos x + \cos (2x)) + (-\cos (2x)+ \cos (3x)) + \cdots,$$

which diverges for every $x\in [-1,1]\setminus \{0\}.$

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