# Joan Lindsay Orr



# Fourier Notes

## Fourier Series

Let $$f\in L^1(0,1)$$ be a Riemann integrable function on $$[0, 1]$$ and define the Fourier coefficients by $\hat{f}(n) = \int_0^1 f(x) e^{-2\pi i nx} \,dx \qquad n\in\ZZ$

Define the $$N$$'th partial sum of the Fourier series for $$f$$ to be: $S_Nf(x) = \sum_{n=-N}^N \hat{f}(n) e^{2\pi inx}$

Suppose that $$f$$ is continuous on $$[0,1]$$ and $$\hat{f}(n)$$ is absolutely summable. Then $$S_Nf(x)\rightarrow f(x)$$ uniformly on $$[0,1]$$.

Suppose $$f$$ is Riemann integrable on $$[0, 1]$$ and $$\hat{f}(n) = O(1/|n|)$$.

1. If $$f$$ is continuous at $$x_0\in[0,1]$$ then $$S_Nf(x_0)\rightarrow f(x_0)$$
2. If $$f$$ is continuous on $$[0,1]$$ then $$S_Nf(x)\rightarrow f(x)$$ uniformly on $$[0,1]$$.

The following two results are useful corollaries of the last theorem:

Suppose that $$f$$ is Riemann integrable on $$[0, 1]$$ and $$\hat{f}(n) = O(1/|n|)$$. Then whenever $$f(x-)$$ and $$f(x+)$$ exist, $S_Nf(x_0)\rightarrow \frac{f(x+) + f(x-)}{2}$

Suppose that:

1. $$f$$ is continuously differentiable except at finitely many points in $$[0,1]$$,
2. $$f'(x)$$ is bounded on the set of points at which it exists.

Then, $$f(x+)$$ and $$f(x-)$$ exist for every $$x\in[0,1]$$ and $S_Nf(x) \rightarrow \frac{f(x+) + f(x-)}{2}$ for all $$x$$.

(Carleson) Let $$f\in L^p(0, 1)$$ for $$1 < p \le \infty$$. Then $$S_Nf(x)$$ converges pointwise almost everywhere to $$f(x)$$.

(Fejér) Define the Fejér sum of $$f$$ to be $\sigma_Nf(x) = \frac{1}{N+1}\sum_{n=0} ^ N S_nf(x) = \sum_{n=-N}^N \frac{N + 1 - |n|}{N+1} \hat{f}(n) e^{2\pi i nt}$ Then,

1. If $$f$$ is continuous at $$x_0$$ then $$\sigma_Nf(x_0) \rightarrow f(x_0)$$, and
2. If $$f$$ is continuous on $$[0,1]$$ then $$\sigma_Nf(x) \rightarrow f(x)$$ uniformly on $$[0,1]$$.

(Parseval's Theorem) Let $$f \in L^2(0,1)$$. Then its Fourier coefficients are in $$\ell^2(\ZZ)$$ and $\int_0^1 |f(x)|^2 dx = \sum_{n=-\infty}^\infty |\hat{f}(n)|^2$ Thus the map $$f \in L^2(0,1) \rightarrow {f}\in\ell^2(\ZZ)$$ is unitary.

## Fourier Transforms

Let $$f\in L^1(\RR)$$ and define the Fourier transform $(\F f)(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i x\xi} \,dx$ Likewise for integrable $$g(\xi)$$ define the inverse Fourier transform $(\F^{-1}g)(x) = \int_{-\infty}^\infty g(\xi) e^{2\pi i x\xi} \, d\xi$

(Fourier Inversion Theorem) If $$f \in L^1(\RR)$$ and also $$\F f \in L^1(\RR)$$ then $$(\F^{-1}\F f)(x) = f(x)$$ for almost every $$x\in\RR$$.

If, in addition, $$f$$ is continuous, then $$(\F^{-1}\F f)(x) = f(x)$$ for all $$x\in\RR$$.

(Plancherel Theorem) If $$f\in L^1(\RR)\cap L^2(\RR)$$ then $$\F f\in L^2(\RR)$$ and $\int_{-\infty}^\infty |(\F f)(\xi)|^2\,d\xi = \int_{-\infty}^\infty |f(x)|^2 \,dx$

Thus the map $$f\rightarrow\F f$$ is an isometry from $$f\in L^1(\RR)\cap L^2(\RR)$$ into $$L^2(\RR)$$, which extends to a unitary $$L^2(\RR)\rightarrow L^2(\RR)$$. This unitary is called the Plancherel transform.

The fact that $$\F$$ is unitary means that $$\ip{\F f}{\F g} = \ip{f}{g}$$ and so for $$f, g \in L^2(\RR)\rightarrow L^2(\RR)$$, \begin{align} \ip{\F f}{\F g} &= \int \overline{ f(x)e^{-2\pi i x \xi}} g(x') e^{-2\pi i x' \xi} dx dx' d\xi \\ & = \int \overline{f(x)}g(x') e^{2\pi i (x-x')\xi} dx dx' d\xi \\ &= \ip{f}{g} \\ &= \int \overline{f(x)}g(x) dx \end{align} This justifies use of the shortcut formula $\int e^{2\pi i (x-x')\xi} d\xi = \delta(x-x')$