Fourier Analysis and Introduction Chapter 3 15 c

Let $ f $ be $ 2 pi $-periodic and integrable Rienmann on $[-pi,pi]$.
$ hat f (n) $ is the cofficient fourier.

a) Show that
$$ hat f (n) = – frac {1} {2 pi} int _ {- pi} ^ { pi} f (x + pi / n) e ^ {- inx} $ dx $
Therefore
$$ hat f (n) = frac {1} {4 pi} int _ {- pi} ^ { pi}[f(x)-f(x+pi/n)]e ^ {- inx} dx. $$

(b) Suppose now that $ f $ satisfies a condition of order of the holder $ alpha $to know
$$ | f (x + h) -f (x) | the C | h | ^ {| alpha |} $$
for some people $ 0 < alpha the $ 1, some $ C> $ 0, and all $ x, h $. Use part a) to show
$$ hat f (n) = O ( frac {1} {| n | ^ { alpha}} $$

(c) Prove that the above result can not be improved by showing that the function
$$ f (x) = sum_ {k = 0} ^ { infty} 2 ^ {- k alpha} e ^ {i2 ^ {k} x}, $$
or $ 0 < alpha <1, $ satisfied
$$ | f (x + h) -f (x) | the C | h | ^ { alpha}, $$
and $ hat f (N) = 1 / N ^ { alpha} $ does not matter when $ N = 2 ^ {k}. $

[Tip:To(c)décomposezlasommecommesuit[Hint:For(c)breakupthesumasfollows[Astuce:Pour(c)décomposezlasommecommesuit[Hint:For(c)breakupthesumasfollows$ f (x + h) -f (x) = sum_ {2 ^ {k} le1 / | h |} + sum_ {2 ^ {k}> 1 / | h |}. $To estimate the first sum, use the fact that $ | 1-e ^ {i theta} | <| theta | $ does not matter when $ theta $ is small. To estimate the second sum, use the obvious inequality $ | e ^ {ix} -e ^ {iy} | $ 2]

I proved a) and b), but when I got to c), I was stuck with proving.
There is what I thought:
$$ | f (x + h) -f (x) | = | sum_ {k = 0} ^ { infty} 2 ^ {- k alpha} e ^ {i2 ^ {k} (x + h)} – sum_ {k = 0} ^ { infty} 2 ^ { – k alpha} e ^ {i2 ^ {k} x} | = | sum_ {k = 0} ^ { infty} 2 ^ {-k alpha} e ^ {i2 ^ {k} x} (e ^ {i2 ^ {k} h} -1) | $$

divide that into $ f (x + h) -f (x) = sum_ {2 ^ {k} le1 / | h |} + sum_ {2 ^ {k}> 1 / | h | and use the index:
$$ | sum_ {k = 0} ^ { infty} 2 ^ {- k alpha} e ^ {i2 ^ {k} x} (e ^ {i2 ^ {k} h} -1) | le | sum_ {2 ^ {k} le1 / | h |} 2 ^ {- k alpha} e ^ {i2 ^ {k} x} 2 ^ {k} h | + | sum_ {2 ^ {k}> 1 / | h |} 2 ^ {- k alpha} e ^ {i2 ^ {k} x} 2 | $$

$$ le sum_ {2 ^ {k} le1 / | h |} 2 ^ {- k alpha} + 2 sum_ {2 ^ {k}> 1 / | h |} 2 ^ {- k alpha} $$
$$ le sum_ {2 ^ {k} le1 / | h |} | h | ^ { alpha} +2 sum_ {2 ^ {k}> 1 / | h |} | h | ^ { alpha} $$
But I do not know what the next step is and how does it prove that the result in b) can not be improved.