- #1

Wingeer

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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(k-n)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(y-x)}{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)| = \left|f(x) - \alpha_n ^{-1} \int_{-\pi}^{\pi} f(y) \left(\frac{1+cos(y-x)}{2}\right)^n dy \right| = \left|\alpha_n ^{-1}\int_{-\pi}^{\pi} (f(x)-f(y)) \left(\frac{1+cos(y-x)}{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]|x-y| < 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(y-x)}{2}\right)^n \right| dy[/tex]

Now he chooses to split the integral by using the limits, so that we get:

[tex]\int_{|x-y|<2\delta}\alpha_n ^{-1} \left|(f(x)-f(y)) \left(\frac{1+cos(y-x)}{2}\right)^n \right| dy + \int_{|x-y| \geq 2\delta} \alpha_n ^{-1} \left|(f(x)-f(y)) \left(\frac{1+cos(y-x)}{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_{|x-y| \geq 2\delta} \left|(f(x)-f(y)) \left(\frac{1+cos(y-x)}{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.