Let's consider the , then we know that the inner product is of the form $$\langle f, g\rangle = \frac{1}{2\pi} \int_0^{2\pi} f(\theta) \overline{g(\theta)} , d\theta$$and the norm $$|f|^2 = \frac{1}{2\pi} \int_0^{2\pi}|f(\theta)|^2, d\theta$$
For each integer , let , and observe that that the family is orthonormal; that is $$\langle e_n, e_m\rangle = \delta_{n,m}$$with being the Kronecker delta.
Let be integrable function on the circle and let denote its Fourier coefficients. We see that the Fourier coefficients are represented by inner products of with the elements in the orthonormal set : $$\langle f, e_n\rangle = \frac{1}{2\pi} \int_0^{2\pi} f(\theta) e^{-in\theta}\ d\theta = a_n$$
We see that . Then the orthonormal property of the family and the fact that imply that is orthogonal to , for all . There for we have that $$(f- S_N(f)) \perp \sum_{|n| \le N} b_n e_n$$for any complex numbers . From this fact we get that, , and $$|f|^2 = |f-S_N(f)|^2+\left|\sum_{|n| \le N}a_n e_n\right|^2$$by the orthonormality we have that $$\left|\sum_{|n| \le N}a_n e_n\right|^2 = \sum_{|n| \le N}|a_n|^2$$, we deduce that $$|f|^2 = |f-S_N(f)|^2+\sum_{|n| \le N}|a_n|^2$$ Best Approximation: If is integrable on the circle with Fourier coefficients , then $$|f-S_N(f)| \le \left|f-\sum_{|n|\le N} c_n e_n\right|$$for any complex numbers . Moreover, equality holds precisely when for all .
Th: Let be an integrable function on the circle with . Then we have that:
Mean-square convergence of the Fourier series: $$\frac{1}{2\pi} \int_0^{2\pi} |f(\theta)- S_N(f)(\theta)|^2 , \theta \to 0 \qquad N\to \infty$$or in
Parseval's identity $$\sum_{n = -\infty}^\infty |a_n|^2 = \frac{1}{2\pi} \int_0^{2\pi}|f(\theta)|^2, d\theta$$In general, we can make be periodic then get $$\sum_{n = -\infty}^\infty |a_n|^2 = \frac{1}{L} \int_0^{L}|f(\theta)|^2, d\theta$$ Bessel's Inequality: Let is any orthogonal family of function on the circle, and , then we may deduce that $$\sum_{n = -\infty}^\infty |a_n|^2 \le |f|^2$$
We have associated every element of with a sequence , and Parseval's identity guarantees that .
Riemann-Lebesgue Lemma If is integrable on the circle, then as . An equivalent reformulation, if that if is integrable in , then $$\int_0^{2\pi} f(\theta) \sin(N\theta), d\theta \to 0 \qquad N \to \infty$$and $$\int_0^{2\pi} f(\theta) \cos(N\theta), d\theta \to 0 \qquad N \to \infty$$ Cor: Let be a periodic and of class , then $$\hat f(n) = o(1/|n|^k)$$
Lemma: suppose that and are integrable on the circle with $$F \sim \sum a_m e^{in\theta} \quad \text{and}\quad G \sim \sum b_n e^{in\theta}$$
Then $$\frac{1}{2\pi} \int_0^{2\pi} F(\theta) \overline{ G(\theta)}, d\theta = \sum_{n = -\infty}^\infty a_n \overline{b_n}$$
and if we a make and be periodic, we get that $$\frac{1}{L} \int_0^L F(\theta) \overline{ G(\theta)}, d\theta = \sum_{n = -\infty}^\infty a_n \overline{b_n}$$
With this we have more things to prove the absolute convergence and hence uniform convergence.
We say that satisfies a Hölder condition of order , namely $$|f(x)-f(y)| \le C|x-y|^\alpha $$for some , and some and all . Then $$\hat f(n) = O(1/|n|^\alpha)$$We note that for , then is Lipschitz continuous.
Bernstein Theorem: If satisfies a Hölder condition of order , then the Fourier series of converges absolutely.
Pointwise Convergence
Th: Let be an integrable function on the circle which is differentiable at a point . Then as .
We can weaken further if we only assume that satisfies a Lipschitz condition at ; that is $$|f(\theta) - f(\theta_0) \le |\theta - \theta_0|$$For some and all .
Riemann Localisation Theorem: Suppose and are two integrable functions defined on the circle, and for some there exists an open interval containing such that $$f|_I = g|_I$$then as .
We also have that near a discontinuity, we have that the Fourier series tend to overshoot/undershoot, by around 9% of the jump at this is called Gibb's phenomenon