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Source: National Board for Higher Mathematics Model Questions, PhD Jan 2018 entrance exam paper, Q2.7 (Direct PDF link, answer key link)

Let $f : \left [-\pi, \pi \right ] \longrightarrow \Bbb R$ be a continuous $2\pi$-periodic function whose Fourier series is given by $$\dfrac {a_0} {2} + \sum\limits_{k=1}^{\infty} \left (a_k \cos kt + b_k \sin kt \right ).$$ Let, for each $n \in \Bbb N,$ $$f_n (t) = \dfrac {a_0} {2} + \sum\limits_{k=1}^{n} \left (a_k \cos kt + b_k \sin kt \right ),$$ and $f_0$ denote the constant function $\dfrac {a_0} {2}.$ Which of the following statements are true?

a. $f_n \to f$ uniformly on $\left [-\pi, \pi \right ].$

b. If $\sigma_n = \dfrac {f_0 + f_1 + \cdots + f_n} {n + 1},$ then $\sigma_n \to f$ uniformly on $\left [-\pi, \pi \right ].$

c. $\displaystyle {\int_{-\pi}^{\pi} {\left \lvert f_n (x) - f(x) \right \rvert}^2\ dx \to 0,}$ as $n \to \infty.$

If $f'$ is piecewise smooth $2\pi$-periodic function then option a is true. What will happen if $f'$ is not given to be piecewise smooth? Is there any counter-example? Also I don't know anything about the last two options.

How do I proceed? Any help will be highly appreciated.

Thanks in advance.

Calvin Khor
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Anacardium
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  • If $f'$ is piecewise smooth $2\pi$-periodic function then option $(a)$ is true. What will happen if $f'$ is not given to be piecewise smooth? Is there any counter-example? Also I don't know anything about the last two options. – Anacardium Oct 11 '20 at 06:22

2 Answers2

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Both b and c are correct.

Option a is not correct. There are continuous functions whose fourier series do not converge pointwise. Regarding your comment on $f'$ being piecewise smooth being enough, actually you only need a small amount of Hölder regularity say $f\in C^\alpha$ for some $\alpha\in(0,1)$ to get uniform convergence.

For b, this follows from the fact that Fejer kernels are "good kernels", and so you have an "approximation to the identity" (these are Googling keywords). One place to find the proof of b is here (random Google result.)

For c, this follows because continuous functions on $[-\pi,\pi]$ (periodic or not) are in particular $L^2([-\pi,\pi])$ functions, and here, the exponential functions are a complete orthonormal set / Schauder basis. Thus the partial expansion converges in $L^2$ norm, and to the original function $f$, which is exactly the statement of c. One can look here for a proof. (There is the related and harder result called Carleson's theorem which says that Fourier series of $L^2$ functions even converges almost everywhere.)

If you prefer a book I would imagine these are covered (at a sufficiently 'elementary' level) in one of Stein and Shakarchi's books.

Calvin Khor
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  • Which book of Stein and Shakarchi are you referring to here? Can you please tell me the corresponding chapter? I'm not familiar with such things (i.e. Fejer kernels, Holder's regularity, Carleson's theorem and so on) as I haven't read those things before. – Anacardium Oct 11 '20 at 07:25
  • @Anacardium I can't tell you the precise page, my books are not with me. But there should be one called Fourier Analysis if I recall correctly, so that one. – Calvin Khor Oct 11 '20 at 07:26
  • @Anacardium I will add, if you haven't heard of Hölder regularity, then this question maybe is a little too advanced, so have some patience when learning the material – Calvin Khor Oct 11 '20 at 07:33
  • The pdf (provided by you) regarding $L^2$-convergence of Fourier series contains a notation $L^2 (\Bbb T).$ What is the meaning of that? Here's the pdf. – Anacardium Oct 11 '20 at 15:41
  • @Anacardium $\Bbb T$ is the circle represented as the quotient group $\mathbb R/\mathbb Z$ – Calvin Khor Oct 12 '20 at 01:18
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I think choice c is correct which is basically the whole motivation behind the study of Fourier series. They approximate a function using sinusoids.

LogicalGooGoo
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