Is a Function Zero if All Its Fourier Coefficients are Zero?

[boombaby]

Homework Statement

Suppose that f is an integrable function (and suppose it's real valued) on the circle with c_n=0 for all n, where c_n stands for the coefficient of Fourier series. Then f(p)=0 whenever f is continuous at the point p.

Homework Equations

The Attempt at a Solution

assuming f is continuous at p=0, and supposing f(p)>0, the book (princeton lectures in analysis I) constructed trigonometric polynomials $$p_{k}(x)=(\epsilon+cos(x))^{k}$$ so that $$\int_{-\pi}^{\pi} f(x) p_{k}(x) dx$$ approaches infinity as k approaches infinity, contradicting the fact that the integral should* be zero. I have no idea why it should be zero. I thought c_n is defined to be $$\int_{-\pi}^{\pi} f(x) e^{-inx} dx$$ ? I do not see they are equivalent in some obvious way...what's the idea in it?
Thanks a lot

 

[mighty2000]

for the help.

Hello,

Thank you for your post. I am a scientist specializing in mathematical analysis and I would be happy to provide some insight into this problem.

Firstly, let's define some terms to make sure we are on the same page. An integrable function is one that is Lebesgue integrable, meaning its integral is well-defined and finite. The Fourier series of a function f is given by the formula c_n = (1/2π) ∫_0^2π f(x)e^(-inx) dx, where c_n is the n-th coefficient of the series. Now, let's consider the statement that if f is integrable and c_n = 0 for all n, then f(p) = 0 whenever f is continuous at the point p.

To understand this, we first need to consider the concept of convergence in the context of Fourier series. The Fourier series of a function f is said to converge to f if the partial sums of the series, S_n(x) = ∑_k= -n^n c_k e^ikx, converge to f(x) as n approaches infinity. This is known as the pointwise convergence of the Fourier series. However, it is possible for a function to have a different value at a point than the limit of its partial sums, even if the series converges pointwise. This is known as a point of discontinuity for the Fourier series.

Now, in the case where c_n = 0 for all n, this means that the Fourier series of f is identically equal to 0. This also means that the partial sums S_n(x) are equal to 0 for all x, and therefore converge to 0 for all x. This implies that the Fourier series of f converges to 0 pointwise, and therefore f(x) = 0 for all x.

To understand why the integral should be 0, we can use the fact that the Fourier coefficients c_n can be expressed in terms of the integral of f. This is given by the formula c_n = (1/2π) ∫_0^2π f(x)e^(-inx) dx. Now, if c_n = 0 for all n, this means that the integral of f multiplied by e^(-inx) is equal to 0 for all n. This implies that the integral of f multiplied by any trigonometric polynomial p_k(x) = (