Lagrange publicly ridiculed Fourier, and said that only continuous functions can be represented by Equation 1 indeed he proved that Equation 1 holds for continuous-time functions.
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However, we know now that the real truth lies in between Fourier and Lagrange's positions. Take the two functions graphed below for example:. These conditions are known as the Dirichlet Conditions. For each complex a and each non-negative integer i , the identity.
We use induction over n. Using identity 3. The notation [ x ] to denote the greatest integer which does not exceed x will be adopted. As a consequence,. The Fourier—Bessel series on a q -linear grid associated with f is defined as the sum. The last equality in formula 4. Theorem B assures mean convergence of the series 4. With a view to studying pointwise convergence of the series 4. There is no real novelty in this section and we are aware that the pointwise convergence follows from the mean convergence by using known results from linear analysis. However, we believe that the reader may benefit from the following elegant self contained argument, which has been gently provided to us by Professor Juan Arias de Reyna.
In the following lemma it is shown that this convergence also holds pointwise.
Let a k be an arbitrary element of N. In fact this problem leads to the celebrated Carleson Theorem [ 28 ]; for a tutorial exposition see [ 16 ]. Since this orthogonal system is a complete system see, e. This answers a question posed in the concluding section of [ 23 ]. In [ 12 ] a rigorous theory of q -Sturm—Liouville systems was developed.
In particular, it was shown that the set of all normalized eigenfunctions forms an orthonormal basis for L q 2 [ 0 , a ]. Consider the basic difference relation 2. Although n and j are nonnegative integers, relation 6. Moreover, it follows from 6. The first identity of 6. Replacing n by n - 2 in 6. Iterating n - j times the same argument provides the identity. The following result uses 6. The proof, once again carried out by induction, is a bit long and technical.
We will simply present a sketch. Let us now assume that. One must show that there exists C , independent of k and n , such that. This suggests dividing the proof in two cases, according to the behavior of n - k. Now, by definition,. Hence, by 6. Then n - k is bounded below. We provide here a more direct proof, based only on the definition of the Hahn—Exton q -Bessel function and its derivative.
We will present only the main steps of the proof. Thus, if p is a positive integer,.
Since this estimate is independent of p , one can resort to 2. Identity The lemma now follows from 6.
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- Pointwise convergence of vector valued Fourier series - Technische Informationsbibliothek (TIB)?
In [ 25 ] the following upper bound for basic Fourier—Bessel coefficient 4. The proof of the main result depends on a refinement of the above estimates. We start proving the following identity. From 3. We will now rewrite the integrals on the right hand side of 6. Using 3. The evaluation of the second q -integral on the right side of 6. Finally, the lemma follows by substituting 6. To assure the convergence of all q -integrals involved in the previous lemma, one is required to assume the same sufficient conditions of Theorem 1.
We are now in conditions to prove our main result. The beef of the proof consists of finding an n -independent convergent upper bound for the q -Fourier—Bessel series. Notice also that the extra condition involving the point q - 1 can be removed or neglected since it only affects the choice of the constant W. Introducing inequalities 6. Thus, the absolute value of the k th term in 6.
To deal with the first term in 6. From 6. Since the constants A and B are positive and independent of k and n , we finally conclude that. This proves the uniform convergence of the basic Fourier—Bessel series 4. We conclude with some explicit examples of uniformly convergent Fourier—Bessel series on a q -linear grid. Using the q -binomial Theorem [ 31 , 1. Therefore, 4. Thus, we can apply Theorem 1. We would like to thank the unknown referee for the valuable comments and suggestions.
The authors are very grateful to Prof. Juan Arias for pointing out the Lemma 5. Also, stimulating discussions with Prof. The research of L. The research of R.
On The Pointwise Convergence Of Fourier Series by Charles J. Mozzochi
The research of J. Abreu, Email: ta. Cardoso, Email: tp. National Center for Biotechnology Information , U. The Ramanujan Journal.
Ramanujan J. Published online Sep Abreu , 1 R. Cardoso 3. Cardoso 3 Dep. Author information Article notes Copyright and License information Disclaimer. So this is an if and only if condition. If 2 in the exponents above is replaced with some p , the question becomes much harder.
The original proof uses properties of holomorphic functions and Hardy spaces , and another proof, due to Salomon Bochner relies upon the Riesz—Thorin interpolation theorem. The construction of an example of divergence in L 1 was first done by Andrey Kolmogorov see below. For infinity, the result is a corollary of the uniform boundedness principle.
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The problem whether the Fourier series of any continuous function converges almost everywhere was posed by Nikolai Lusin in the s. It was resolved positively in by Lennart Carleson. His result, now known as Carleson's theorem , tells the Fourier expansion of any function in L 2 converges almost everywhere.
Contrariwise, Andrey Kolmogorov , as a student at the age of 19, in his very first scientific work, constructed an example of a function in L 1 whose Fourier series diverges almost everywhere later improved to diverge everywhere. Does the sequence 0,1,0,1,0,1, This does not seem like a very unreasonable generalization of the notion of convergence.
Hence we define. This implies much better convergence properties. Results about summability can also imply results about regular convergence. See Big O notation for the notation O 1. The fact that for some constant c we have. This estimate entails quantitative versions of some of the previous results. For any continuous function f and any t one has.