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 ?
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 $x\in [-1,1]\setminus \{0\}.$