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I need help with the last part of the following problem:
Let f(x) be a 2\pi-periodic and Riemann integrable on [-\pi,\pi].
(a) Assuming f(x) satisfies the Hölder condition of order \alpha
Show that
(b) Proove that the above result cannot be improved by showing that the function
I have done (a). And I was able to show that the sum satisfies the condition. What I don't understand is:
how can I use it to show that the result in (a) cannot be improved?
I thought that the sum can be iterpreted as the Fourier series of g(x), which would say that the Fourier coefficient of g(x) is 1/{n^\alpha}, where n=2^k. That is the resemblence I see between (a) and (b). But I don't know what to do with that.
Let f(x) be a 2\pi-periodic and Riemann integrable on [-\pi,\pi].
(a) Assuming f(x) satisfies the Hölder condition of order \alpha
\left| {f\left( {x + h} \right) - f\left( x \right)} \right| \le C\left| h \right|^\alpha ,
for some 0 < \alpha \le 1, some C > 0 and all x, h.Show that
{\hat f\left( n \right)} \le O ( \frac {1} { \left| n \right| ^ \alpha} )
where \hat f\left( n \right) is the Fourier koefficient.(b) Proove that the above result cannot be improved by showing that the function
g\left( x \right) = \sum\limits_{k = 0}^\infty {2^{ - k\alpha } e^{i2^k x} }
also with 0 < \alpha \le 1, satisfies\left| {g\left( {x + h} \right) - g\left( x \right)} \right| \le C\left| h \right|^\alpha .
I have done (a). And I was able to show that the sum satisfies the condition. What I don't understand is:
how can I use it to show that the result in (a) cannot be improved?
I thought that the sum can be iterpreted as the Fourier series of g(x), which would say that the Fourier coefficient of g(x) is 1/{n^\alpha}, where n=2^k. That is the resemblence I see between (a) and (b). But I don't know what to do with that.
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