Fourier Series - proving function is continuous

In summary, the Fourier coefficients of an integrable, periodic function f must satisfy \sum_{-\infty}^{\infty} n^6 |\hat{f}(n)|^2 < \infty in order for f to be continuous. This can be proven using either the uniform convergence of a sequence of continuous functions, or the Riemann-Lebesgue lemma.
  • #1
ramdayal9
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Homework Statement


Let f be an integrable, periodic function whose Fourier coefficients satisfy [itex] \sum_{-\infty}^{\infty} n^6 |\hat{f}(n)|^2 < \infty [/itex]. Prove that f is continuous.


Homework Equations



Looking at my notes, the only relevant things i have for this question (i think) are Bessel's inequality (but that requires f to be continuous...), the Riemann-Lebesgue Theorem, Parseval's equality and also the decay and regularity of Fourier series. The statement that the uniform limit of a sequence of continuous functions is continuous, maybe I can use this?


The Attempt at a Solution



I thought this question would be simple, but I'm finding that its not. I have tried 2 methods:
1) [itex]\sum_{-\infty}^{\infty} n^6 |\hat{f}(n)|^2 < \infty [/itex] implies [itex] n^6 |\hat{f}(n)|^2 \rightarrow 0 [/itex] hence [itex]|\hat{f}(n)| \rightarrow 0 [/itex]. I need to show that f is continuous i.e. the converse of the Riemann Lebesgue Lemma. How would I proceed from here?
2) using L^2 convergence theorem (parseval's), [itex]\sum_{-\infty}^{\infty} |c_n|^2 \< \infty [/itex] hence there exists function g such that [itex]\hat{g}(n)=c_n=n^3 |\hat{f}(n)| [/itex]. Is this the route I should proceed?

The hypothesis also states that f is integrable, but i don't know how to use this here! I am also thinking of the M test, but don't know if I can apply it to a sum like this... Any help will be appreciated!
 
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  • #2




Thank you for posting this interesting problem. I would approach this question by first looking at the properties of the Fourier coefficients and how they relate to the continuity of a function. In particular, the fact that \sum_{-\infty}^{\infty} n^6 |\hat{f}(n)|^2 < \infty tells us that the Fourier coefficients decay very quickly as n increases. This suggests that the function f is very smooth and regular, and therefore likely to be continuous.

To prove this, we can use the fact that the uniform limit of a sequence of continuous functions is continuous. As you mentioned, we can construct a sequence of functions g_n such that \hat{g_n}(n) = n^3 |\hat{f}(n)|, and then use Parseval's equality to show that this sequence converges uniformly to f. Since each g_n is continuous, and the uniform limit of continuous functions is continuous, we can conclude that f is continuous.

Alternatively, we can also use the Riemann-Lebesgue lemma to show that the Fourier coefficients of f must decay at infinity. This, combined with the fact that they also decay very quickly, implies that f must be continuous.

I hope this helps guide you towards a solution. Good luck!
 

Related to Fourier Series - proving function is continuous

1. What is a Fourier Series?

A Fourier Series is a mathematical representation of a periodic function as a sum of sinusoidal functions. It is used to analyze and approximate complex functions that are difficult to solve using traditional methods.

2. How is a function proved to be continuous using Fourier Series?

A function can be proved to be continuous using Fourier Series by showing that the series converges to the same value as the function at each point. This is known as the pointwise convergence of the series.

3. What is the importance of proving a function to be continuous using Fourier Series?

Proving a function to be continuous using Fourier Series is important because it allows for the accurate approximation of complex functions, which has applications in various fields such as engineering, physics, and signal processing.

4. Can any function be represented by a Fourier Series?

No, not all functions can be represented by a Fourier Series. The function must be periodic and have a finite number of discontinuities in order to be represented accurately by a Fourier Series.

5. What is the difference between Fourier Series and Fourier Transform?

Fourier Series is used to approximate periodic functions, while Fourier Transform is used to analyze non-periodic functions. Fourier Transform also considers the entire domain of a function, while Fourier Series only considers a single period.

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