
#1
Jan1613, 10:32 AM

P: 79

Hi.
Just going through my notes from the last lecture I remember having some troubles understanding the proof the lecturer gave for the following theorem: Suppose that f is Riemann integrable and that all its Fourier coefficients are equal to 0, then f(x)=0 at all points of continuity. The proof is a bit tricky so I will sketch the basic gists of it. We start by defining [tex]\delta_n = \frac{\left(\frac{1+cos(x)}{2}\right)^n}{\alpha_n}[/tex] where [tex]\alpha_n = \int_{\pi}^{\pi} \left(\frac{1+cos(x)}{2}\right)^n dx[/tex]. This implies of course that [tex]\int_{\pi}^{\pi} \delta_n dx = 1, \forall n[/tex] He then proceeds by showing that [tex]\frac{1}{\alpha_n} \leq \left(2\delta \left(\frac{1+cos(\delta)}{2}\right)^n\right)^{1}[/tex] Which will be useful later in the proof (this part was also understandable). Now: [tex]\left(\frac{1+cos(x)}{2}\right)^n = \left(\frac{exp(\frac{ix}{2}) + exp(\frac{ix}{2})}{2}\right)^{2n}[/tex] He then uses the binomial theorem on this expression to get: [tex]\sum_{k=0}^{2n} {2n \choose k} e^{i(kn)x}[/tex] My first question is then: Where did the 2's from the denominator go? Now he uses this expansion to conclude that: [tex]\int_{\pi}^{\pi} f(y) \left(\frac{1+cos(yx)}{2}\right)^n dy = 0[/tex] By arguing that the Fourier coefficients are 0 by assumption. My second question is regarding this step. I do not understand it at all. any clarity at all will be a tremendous help. Then [tex]f(x) = \leftf(x)  \alpha_n ^{1} \int_{\pi}^{\pi} f(y) \left(\frac{1+cos(yx)}{2}\right)^n dy \right = \left\alpha_n ^{1}\int_{\pi}^{\pi} (f(x)f(y)) \left(\frac{1+cos(yx)}{2}\right)^n dy \right[/tex] Which follows from how the delta function is defined. He then proceeds by noting that [tex]\lim_{y \to x}f(y) = f(x)[/tex] so that for any given epsilon>0 there exists 2delta such that [tex]xy < 2\delta \Leftarrow f(x)  f(y) < \epsilon[/tex] By definition of limits. Now by the triangle inequality for integrals we have that the expression over is less than or equal to: [tex]\alpha_n ^{1}\int_{\pi}^{\pi} \left(f(x)f(y)) \left(\frac{1+cos(yx)}{2}\right)^n \right dy[/tex] Now he chooses to split the integral by using the limits, so that we get: [tex]\int_{xy<2\delta}\alpha_n ^{1} \left(f(x)f(y)) \left(\frac{1+cos(yx)}{2}\right)^n \right dy + \int_{xy \geq 2\delta} \alpha_n ^{1} \left(f(x)f(y)) \left(\frac{1+cos(yx)}{2}\right)^n \right dy[/tex] Now the first integral is smaller than epsilon due to the limits and also by the normalisation of delta. In the second integral we pull out alpha and use the inequality mentioned earlier. Now he also claims that: [tex]\int_{xy \geq 2\delta} \left(f(x)f(y)) \left(\frac{1+cos(yx)}{2}\right)^n \right dy \leq 2M \left(\frac{1+cos(2\delta)}{2}\right)^n[/tex] Which I believe to be the case since f is Riemann integrable and therefore must be bounded by, say M. Also on the right side we have the maximal value that the cosine expression takes, and therefore the inequality is justified. Is this correct thinking? Now by taking limits we get the conclusion that [tex]f(x) < \epsilon[/tex] where epsilon was arbitrarily chosen, and therefore f(x)=0 where it is continuous. I know this result probably goes a bit deeper and can be better understood with measure theory and the like. I have not had any courses with this yet, so my question is then: Is there a more general formulation of this theorem where we do not restrict the functions to be Riemann integrable? Thanks in advance. 



#2
Jan2913, 12:46 AM

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PF Gold
P: 2,890

This is quite a long problem, so I will only address part of it for now. Regarding your first question:
$$\begin{align} \left(\frac{\exp\left(\frac{ix}{2}\right) + \exp\left(\frac{ix}{2}\right)}{2}\right)^{2n} &= \frac{1}{2^{2n}}\sum_{k=0}^{2n} {2n \choose k} \left(\exp\left(\frac{ix}{2}\right)\right)^{k} \left(\exp\left(\frac{ix}{2}\right)\right)^{2nk} \\ &= \frac{1}{2^{2n}}\sum_{k=0}^{2n} {2n \choose k} \exp\left(\frac{ixk}{2}\right) \exp\left(\frac{ix(2nk)}{2}\right) \\ &= \frac{1}{2^{2n}}\sum_{k=0}^{2n} {2n \choose k} \exp\left(\frac{ixk}{2}\right) \exp\left(\frac{ixk}{2}\right) \exp\left(\frac{i2xn}{2}\right) \\ &= \frac{1}{2^{2n}}\sum_{k=0}^{2n} {2n \choose k} \exp\left(ixk\right) \exp\left(ixn\right) \\ &= \frac{1}{2^{2n}}\sum_{k=0}^{2n} {2n \choose k} \exp\left(ix(kn)\right) \end{align}$$ So this matches your lecturer's expression except for the [itex]1/2^{2n}[/itex] factor, but his conclusion that the result integrates to zero is true with or without this factor: $$\begin{align}\int_{\pi}^{\pi} f(y) \left(\frac{1+cos(yx)}{2}\right)^n dy &= \frac{1}{2^{2n}} \int_{\pi}^{\pi} f(y) \sum_{k=0}^{2n} {2n \choose k} \exp\left(i(yx)(kn)\right) dy \\ &= \frac{1}{2^{2n}} \sum_{k=0}^{2n} {2n \choose k} \int_{\pi}^{\pi} f(y) \exp\left(i(yx)(kn)\right) dy\\ &= \frac{1}{2^{2n}} \sum_{k=0}^{2n} {2n \choose k} \exp(ix(kn)) \int_{\pi}^{\pi} f(y) \exp(iy(kn)) dy \\ \end{align}$$ Note that $$\int_{\pi}^{\pi} f(y) \exp\left(iy(kn)\right) dy$$ is a Fourier coefficient of ##f##, hence zero by assumption. Therefore the whole expression reduces to zero. I will read over the rest of your post and reply again if I can offer any further insight. 



#3
Jan2913, 01:29 AM

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PF Gold
P: 2,890

(a) For all [itex]n \geq 1[/itex], $$ \int_{\pi}^{\pi} K_n(x) dx = 1$$ (b) There exists [itex]M > 0[/itex] such that for all [itex]n \geq 1[/itex], $$\int_{\pi}^{\pi} K_n(x) dx \leq M$$ (c) For every [itex]\delta > 0[/itex], $$\int_{\delta \leq x \leq \pi} K_n(x)dx \rightarrow 0 \textrm{ as } n \rightarrow \infty$$ and they prove some nice theorems regarding what you can do with a good family of kernels, as well as giving several examples of such families. You can check for yourself whether the kernels in your problem satisfy these conditions. 



#4
Feb413, 04:33 PM

P: 79

Basic Fourier analysis proof
Thank you for your answer and time!
We are actually using the book by Stein and Shakarchi in this course. I am not sure if I like it or not; It is certainly something else from what I've read before. Also I am just recently diving into analysis, so I might need to get used to this way of thinking. Could you please elaborate on the purpose of kernels? Especially in how they make limiting arguments more feasible. Is not the integrand supposed to be squared in the last paragraph? Since that is the norm on L2(pi,pi)? Or am I wrong? Probably a difficult question, but how small does the total length have to be? And how do the length compare with the ones of neighbourhoods of open intervals? Another question arose last week when we were talking about the basis of L2(pi,pi). We proved that the family of function e^(inx) for integer n is a basis for the aforementioned space. Then he posed a related question: Prove that $$\frac{\sin(nx)}{\sqrt{\pi / 2}}$$ for integer n is a basis for the space L2(0,pi). First off the family of functions are definitely orthonormal by a quick computation of an integral. The completion part is however stalling me. I really want to use the fact that the exponentials are a basis to prove that this family is a basis, but I cannot really see how to go about that. Maybe a translation and scaling, or something up that alley? I really appreciate you taking your time to help! 



#5
Feb413, 06:02 PM

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PF Gold
P: 2,890

$$f_n(x) = \int_{\pi}^{\pi} f(y)K_n(xy) dy$$ is continuous at all points even if the original function ##f## was not. Furthermore, at any point ##x## where ##f## is continuous, we have ##f_n(x) \rightarrow f(x)##. Even better, if S is the set of points where ##f## is continuous, and S is compact, then ##f_n \rightarrow f## uniformly on S. Now, why is this useful in Fourier analysis? One of the key questions is under what circumstances we can reconstruct ##f## given its Fourier coefficients. The simplest candidate way to do this is to define a sequence of partial sums: $$f_n(x) = \sum_{k=n}^{n} \hat{f}(k) e^{ikx}$$ If you do some clever rearranging, it turns out that this can be written as a convolution with a kernel: $$f_n(x) = \int_{\pi}^{\pi} f(y) D_n(xy) dy$$ where ##D_n(x) = \sin((2n+1)x)/\sin(x)## is the socalled Dirichlet kernel. Unfortunately it turns out that the Dirichlet kernel is not a "good" kernel; it violates the second condition laid out by Stein and Shakarchi because there is no M such that $$\int_{\pi}^{\pi}D_n(x) dx \leq M$$ for all ##n##. A consequence of this is that we aren't guaranteed that the partial sums of the Fourier series will converge to the original function, even at points where that function is continuous. (Indeed, explicit counterexamples have been constructed.) However, there are other ways to reconstruct the function ##f## from its Fourier coefficients. One way is to take averages of the partial sums of the Fourier series. In other words, $$g_n(x) = \frac{1}{n}\sum_{m=1}^{n} f_m(x)$$ where ##f_m(x)## is as defined above. This turns out to be equivalent to forming a sum like $$g_n(x) = \sum_{k=n}^n \hat{f}(k) w_n(k) e^{ikx}$$ where ##w_n(k)## is a sequence of weighting factors that has a triangular profile between n and n. It turns out that this is also equivalent to convolution with a kernel: $$g_n(x) = \int_{\pi}^{\pi} f(y) S_n(xy) dy$$ Here ##S_n(x)## is called a Fejer kernel, which (if I recall correctly) equals the SQUARE of ##D_n(x)## (times a scale factor like 1/n), and this does turn out to be a good kernel. Therefore the sequence of AVERAGES of the partial sums of the Fourier series does converge to the original function at all points of continuity. (By the way, there's no hope at points of discontinuity, since we can change the function's value at each of these without having any effect on its Fourier coefficients, because these are defined by an integral.) 



#6
Feb513, 04:34 AM

P: 79

Wow. Great answer again. Really clarifying, but still leaves me curious. I recently bought the real analysis book by McDonald and Weiss and I really enjoy reading in it.
It is indeed meant to be the open interval, yes. 



#7
Feb513, 03:23 PM

P: 79

I have a suggestion:
Let ##f \in L^2(0, \pi)##, then we have: $$ f' = \left\{ \begin{array}{l} f(x) \text{ if } x \in (0, \pi)\\ f(x) \text{ if } x \in (\pi,0) \end{array} \right.$$ and define f'(0) to be the average of the limiting processes of each of the piecewise functions (or something like that). Then f' is in ##L^2( \pi, \pi)## and it is odd and can therefore be expressed by a linear combination of sin(nx). This was what me and a friend came up with. Now that I see over it myself, I can't help but think: What's stopping us from doing the same argument with even functions? And also, will the claim imply that the only functions in L2(0,pi) is odd? That seems weird ... 



#8
Feb513, 04:30 PM

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HW Helper
PF Gold
P: 2,890

You know that the family ##f_m(x) = e^{imx}## is a basis for ##L^2(\pi, \pi)##, or for any interval of length ##2\pi##, for example ##L^2(0, 2\pi)##. So scale the x axis by a factor of 2 to obtain
$$g_m(x) = f_m(2x) = e^{i2mx}$$ as a basis for ##L^2(0, \pi)##. Now, can you show that each ##g_m## can be expanded in terms of the proposed basis ##\sin(nx)/\sqrt{\pi/2}##? If so, you may be able to argue that ##\sin(nx)/\sqrt{\pi/2}## is indeed a basis for ##L^2(0, \pi)##. Consider the ##n##'th Fourier coefficient of ##g_m## with respect to the proposed basis. I'm too lazy to keep track of constant scale factors so I'll just write ##K## for these: $$\begin{align}c_{n,m} &= K \int_0^{\pi} e^{i2mx} \sin(nx) dx \\ &= \frac{K}{2i} \int_0^{\pi} e^{i2mx} [e^{inx}  e^{inx}] dx \\ &= \frac{K}{2i} \left[\int_0^{\pi} e^{i(2m+n)x} dx  \int_0^{\pi} e^{i(2mn)x} dx \right] \end{align}$$ Note that if ##k## is an integer, then $$\int_0^{\pi} e^{ikx} dx = \pi$$ if ##k## = 0, and otherwise we have $$\int_0^{\pi} e^{ikx} dx = \frac{1}{ik} [e^{ik\pi}  1] = \frac{1}{ik} [(1)^k  1] = \begin{cases} 2i/k & \textrm{if }k \textrm{ is odd} \\ 0 & \textrm{if }k \textrm{ is even} \\ \end{cases}$$ It's clear that we can continue carrying out this calculation to end up with a formula for ##c_{n,m}##, which will give us $$g_m(x) \sim \sum_{n=\infty}^{\infty} c_{n,m} \sin(nx)$$ where I wrote ~ instead of = because convergence needs to be established. Assuming that works out, then an arbitrary function ##f\in L^2(0, \pi)## can be written as $$f(x) \sim \sum_{m=\infty}^{\infty} a_m g_m(x)$$ where ##a_m## is the ##m##'th Fourier coefficient for ##f## in terms of the basis ##g_m##. This then becomes $$f(x) \sim \sum_{m=\infty}^{\infty} a_m\left( \sum_{n=\infty}^{\infty} c_{n,m} \sin(nx)\right)$$ and if the sums are wellbehaved, we can rearrange as follows $$f(x) \sim \sum_{n=\infty}^{\infty} \left(\sum_{m=\infty}^{\infty} a_m c_{n,m} \right) \sin(nx)$$ thereby giving us, in principle, a Fourier coefficient ##d_n## for ##f## in terms of the basis ##\sin(nx)## equal to $$d_n = \sum_{m=\infty}^{\infty} a_m c_{n,m}$$ Of course, all of these manipulations will need to be justified. Absolute convergence will be required in order to move summations around like that. 


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