Uniform convergence of Fourier Series satisfying Lipschitz condition

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SUMMARY

The discussion centers on the uniform convergence of Fourier series for integrable functions satisfying the Lipschitz condition. The participant outlines their struggle with estimating the sum of Fourier coefficients and demonstrates the application of the Cauchy-Schwarz inequality to bound the L² norm of the function. They establish that the series converges absolutely by showing that the Fourier coefficients converge to zero, ultimately solving the problem after initially being lost in the details of the convergence process.

PREREQUISITES
  • Understanding of Fourier series and Fourier coefficients
  • Knowledge of the Lipschitz condition and its implications
  • Familiarity with the Cauchy-Schwarz inequality
  • Basic concepts of absolute convergence in series
NEXT STEPS
  • Study the implications of the Lipschitz condition on Fourier series convergence
  • Learn about the properties of L² spaces and norms
  • Explore advanced techniques for estimating Fourier coefficients
  • Investigate the relationship between absolute convergence and rearrangement of series
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Mathematicians, students studying real analysis, and anyone interested in the convergence properties of Fourier series and their applications in functional analysis.

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Homework Statement



f is integrable on the circle and satisfies the Lipschitz condition (Holder condition with a=1). Show that the series converges absolutely (and thus uniformly). i literally spent about 20 hours on this problem today but i just could not figure it out. i have a feeling it's not that hard, but i am having a hell of a time estimating the sum that i need to show the Fourier coefficients converge absolutely.

Homework Equations


|f(x) - f(y)| <= K |x-y| for all x, y.
I have done up to and including this part of the problem (the easiest part).
show:
\infty
\sum (|sin(nh)|^2) |a(n)|^2 \leq (K^2)*(h^2)
-\infty

now take h = pi/(2^p+1) and show: (i'm going to call this inequality ***)
\sum |a(n)|^2 \leq (K^2)(pi^2)/(2^2p+1)
(2^p-1) < |n| <= 2^(p)

where a(n) is the nth Fourier coefficient. No need to comment on the above.

The Attempt at a Solution


Here's the part i can't get: estimate the partial sums of the Fourier coefficients where the absolute value of the summation index n ranges over 2^(2p-1) to 2^p as above, in order to show that the Fourier series converges absolutely. Hint: Use the Cauchy Schwartz inequality.

I applied the Cauchy Schwartz inequality to the inner product of f and the complex exponential to get a bound for this quantitiy which is the L^2 norm of f. this is ofcourse the inner product of f with itself square root. but the inner product of the partial sums of the Fourier series must converge to the inner product of f with itself, and this inner product is akin to the left side of *** so that the Fourier coefficients must converge to zero. anyway, this is not the answer that i need because i need to estimate the partial sums of the Fourier coefficients in the given range. i tried lots of other stuff too, like moving various sums inside the inner product, but i feel there is something fundamental that i am missing. please give me hints or references but not solutions.

PS: i also tried summing *** over 1 to N so that the left side of *** becomes the partial sums of the inner product of the two sided sequence of Fourier coefficients in the little L over z (sorry don't really know how to say this) norm with itself [i mean the vector space of all two sided sequences whose infinite sum of absolute values of components converges]; the right side becomes a geometric series ofcourse. blah, sorry i am kind of lost in this course.
 
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hey everybody i solved it so don't worry, i will be posting the solution in a couple days here for those who are interested. i was allowing p to vary instead of keeping it fixed. The one thing i wasn't sure about was interchanging a finite sum with the infinite sum of an absolutely convergent series. however i think it's okay because with absolute convergence we can sum in which ever order we want... (i am talking about the inner product in the little L squared Z vector space.)
 

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