real analysis Prove that f is analytic on (a, b) iff f ' is analytic on (a, b)?

Question

Let f ∈ C^∞(a, b). Prove that f is analytic on (a, b) if and
only if f ' is analytic on (a, b).

I need the process, thank you~!

Answer 1

Fix x_0 in (a,b) and suppose f is analytic on (a,b). Then there is a sequence {a_n} such that f(x) = ∑(n = 0 to ∞) a_n (x - x_0)^n within a disc |x - x_0| < R. Without loss of generality, suppose R is the radius of convergence of f. Consider the partial sums



∑(n = 1 to N) na_n(x - x_0)^(n - 1),



obtained by differentiating the the partial sums of f. Since R is the radius of convergence of f, 1/R = lim sup |a_n|^(1/n). Thus



lim sup |na_n|^(1/n) = lim n^(1/n) • lim sup |a_n|^(1/n) = lim sup |a_n|^(1/n) = 1/R.



So ∑(n = 1 to ∞) na_n(x - x_0)^(n - 1) converges with the same radius of convergence R. In particular, since the original series and the differentiated series are uniformly convergent inside the disc |x - x_0| < R, we deduce that



f'(x) = ∑(n = 1 to ∞) na_n(x - x_0)^(n - 1) for |x - x_0| < R.



Since x_0 is arbitrary, f' is analytic on (a,b).



Conversely, suppose f' is analytic on (a,b). Fix x_0 in (a,b) and let {a_n} be a sequence such that f'(x) = ∑(n = 0 to ∞) a_n (x - x_0)^n inside the disc |x - x_0| < R. Again, suppose without loss of generality that R is the radius of convergence of this power series. Consider the partial sums



∑(n = 0 to N) a_n / (n + 1) (x - x_0)^(n + 1)



obtained by integrating the partial sums of f'. Since R is the radius of convergence of f', 1/R = lim sup |a_n|^(1/n), hence



lim sup |a_n / (n + 1)|^(1/n) = lim [1/(n + 1)]^(1/n) lim sup |a_n|^(1/n) = lim sup |a_n|^(1/n) = 1/R. So the series ∑(n = 0 to ∞) a_n / (n + 1) (x - x_0)^(n + 1) converges with the same radius of convergence R. In particular, since the original series and the differentiated series converges uniformly in the disc |x - x_0| < R, we deduce that



f(x) = ∑(n = 0 to ∞) a_n (x - x_0)^n for |x - x_0| < R.



Since x_0 is arbitrary, f is analytic on (a,b).

Answer 2

I'm not 100% sure what kind of proof you want and what the background behind the proof may be.

Anyway:



If f is analytic on (a,b) then for every y in (a,b) f is equal to its Taylor series in some neighborhood of y.

(The Taylor series of course exists, as we are given that f ∈ C^∞(a, b) ).



So, let y be some point in (a,b). Around y, f is given by:



f(x) = f(y) + f'(y) * (x-y) + f''(y)/2! * (x-y)^2 + f(3')(y)/3! * (x-y)^3 + ....



Therefore by differentiating:



f'(x) = f'(y) + f''(y)/2! * 2(x-y) + f(3')(y)/3! * 3(x-y)^2 + f(4')(y)/4! * 4(x-y)^3 + ....

or

f'(x) = f'(y) + (f')''y) * (x-y) + (f')''(y)/2! * (x-y)^2 + (f')(3')(y)/3! * (x-y)^3 + ....



But the second side is exactly the Taylor series for f' and the left side is f' itself



Therefore f' is equal to its Taylor series and is therefore also analytic



For the inverse:

If f' is analytic on (a,b) then for every y in (a,b) f' is equal to its Taylor series in some neighborhood of y.

(The Taylor series of course exists, as we are given that f ∈ C^∞(a, b) and therefore

f' ∈ C^∞(a,b) as well).



So, let y be some point in (a,b). Around y, f' is given by:



f'(x) = f'(y) + f''(y) * (x-y) + f'''(y)/2! * (x-y)^2 + f(4')(y)/3! * (x-y)^3 + ....



Integrating both sides from y to x in (a,b) and considering f is an anti-derivative of f'



f(x) - f(y) =



= f'(y) * (x-y) + f''(y) * 1/2 * (x-y)^2 + f'''(y)/2! * 1/3 * (x-y)^3 + f(4')(y)/3! * 1/4 * (x-y)^4 + ....



or



f(x) = f(y) + f'(y) * (x-y) + f''(y)/2! * (x-y)^2 + f'''(y)/3! * (x-y)^3 + f(4')(y)/4! * (x-y)^4 + ....



Which means f equals its Taylor series and is therefore analytic as well.

Answer 3

f could be non-end for this to be actual. assume f is cts on I = [a,b] everywhere different than in some unspecified time interior the destiny c the place a< c < b. enable M = sup{f(x) : x is on I} m = inf{f(x) : x is on I} restoration eps >0 and choose del < eps / (M - m +a million). now seem a function g- defined as g- = the limit of f to [a, c - del]. this guy is cts and integrable, so there exists a partition P- s.t. S(P-, g-) - s(P-,g-) < eps. that is in lots of cases accomplished likewise for g+ = the limit of f to [c + del, b]. then there exists partition P+ s.t. S(P+, g+) - s(P+,g+) < eps. now enable Q be P- U P+ (the union). Q is a partition of [a,b] and S(Q,f) - s(Q,f) = [S(P-,g-) + M*2del + S(P+,g+)] - [s(P-,g-) + m*2del + s(P+,g+)] = yada yada ... < ...at last ...<4eps.